Answer: 46

Constant acceleration
One-dimensional motion

Prompt: An object starts at 22 m/s and accelerates at 1 m/s2 for 2 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 22·2 + 0.5·1·22 = 46.0.


Answer: 88.5

Constant acceleration
One-dimensional motion

Prompt: An object starts at 25 m/s and accelerates at 3 m/s2 for 3 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 25·3 + 0.5·3·32 = 88.5.


Answer: 112

Constant acceleration
One-dimensional motion

Prompt: An object starts at 9 m/s and accelerates at 2 m/s2 for 7 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 9·7 + 0.5·2·72 = 112.0.


Answer: -14

Constant acceleration
One-dimensional motion

Prompt: An object starts at 5 m/s and accelerates at -2 m/s2 for 7 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 5·7 + 0.5·-2·72 = -14.0.


Answer: 132

Constant acceleration
One-dimensional motion

Prompt: An object starts at 19 m/s and accelerates at 1 m/s2 for 6 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 19·6 + 0.5·1·62 = 132.0.


Answer: 32

Constant acceleration
One-dimensional motion

Prompt: An object starts at 15 m/s and accelerates at 1 m/s2 for 2 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 15·2 + 0.5·1·22 = 32.0.


Answer: 21

Constant acceleration
One-dimensional motion

Prompt: An object starts at 4 m/s and accelerates at 2 m/s2 for 3 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 4·3 + 0.5·2·32 = 21.0.


Answer: 34

Constant acceleration
One-dimensional motion

Prompt: An object starts at 18 m/s and accelerates at -1 m/s2 for 2 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 18·2 + 0.5·-1·22 = 34.0.


Answer: 182

Constant acceleration
One-dimensional motion

Prompt: An object starts at 19 m/s and accelerates at 2 m/s2 for 7 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 19·7 + 0.5·2·72 = 182.0.


Answer: 96

Constant acceleration
One-dimensional motion

Prompt: An object starts at 22 m/s and accelerates at -2 m/s2 for 6 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 22·6 + 0.5·-2·62 = 96.0.


Answer: 100

Constant acceleration
One-dimensional motion

Prompt: An object starts at 15 m/s and accelerates at 2 m/s2 for 5 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 15·5 + 0.5·2·52 = 100.0.


Answer: 256

Constant acceleration
One-dimensional motion

Prompt: An object starts at 20 m/s and accelerates at 3 m/s2 for 8 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 20·8 + 0.5·3·82 = 256.0.


Answer: 63

Constant acceleration
One-dimensional motion

Prompt: An object starts at 2 m/s and accelerates at 2 m/s2 for 7 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 2·7 + 0.5·2·72 = 63.0.


Answer: 84

Constant acceleration
One-dimensional motion

Prompt: An object starts at 15 m/s and accelerates at 3 m/s2 for 4 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 15·4 + 0.5·3·42 = 84.0.


Answer: 112

Constant acceleration
One-dimensional motion

Prompt: An object starts at 6 m/s and accelerates at 2 m/s2 for 8 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 6·8 + 0.5·2·82 = 112.0.


Answer: 26

Constant acceleration
One-dimensional motion

Prompt: An object starts at 12 m/s and accelerates at 1 m/s2 for 2 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 12·2 + 0.5·1·22 = 26.0.


Answer: 64

Constant acceleration
One-dimensional motion

Prompt: An object starts at 14 m/s and accelerates at 1 m/s2 for 4 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 14·4 + 0.5·1·42 = 64.0.


Answer: 44

Constant acceleration
One-dimensional motion

Prompt: An object starts at 13 m/s and accelerates at -1 m/s2 for 4 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 13·4 + 0.5·-1·42 = 44.0.


Answer: -10

Constant acceleration
One-dimensional motion

Prompt: An object starts at 3 m/s and accelerates at -2 m/s2 for 5 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 3·5 + 0.5·-2·52 = -10.0.


Answer: 107.5

Constant acceleration
One-dimensional motion

Prompt: An object starts at 19 m/s and accelerates at 1 m/s2 for 5 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 19·5 + 0.5·1·52 = 107.5.


Answer: 8

Constant acceleration
One-dimensional motion

Prompt: An object starts at 4 m/s and accelerates at -1 m/s2 for 4 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 4·4 + 0.5·-1·42 = 8.0.


Answer: 144

Constant acceleration
One-dimensional motion

Prompt: An object starts at 22 m/s and accelerates at -1 m/s2 for 8 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 22·8 + 0.5·-1·82 = 144.0.


Answer: 34.5

Constant acceleration
One-dimensional motion

Prompt: An object starts at 13 m/s and accelerates at -1 m/s2 for 3 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 13·3 + 0.5·-1·32 = 34.5.


Answer: 50

Constant acceleration
One-dimensional motion

Prompt: An object starts at 24 m/s and accelerates at 1 m/s2 for 2 s. How far does it travel (metres)?

Solution: s = ut + 1⁄2at2 = 24·2 + 0.5·1·22 = 50.0.


Answer: 2704

Prompt: Compute the kinetic energy in joules of a 8 kg mass moving at 26 m/s.

Solution: KE = 1⁄2mv2 = 0.5·8·262 = 2704.0.


Answer: 80

Prompt: Compute the kinetic energy in joules of a 10 kg mass moving at 4 m/s.

Solution: KE = 1⁄2mv2 = 0.5·10·42 = 80.0.


Answer: 3364

Prompt: Compute the kinetic energy in joules of a 8 kg mass moving at 29 m/s.

Solution: KE = 1⁄2mv2 = 0.5·8·292 = 3364.0.


Answer: 392

Prompt: Compute the kinetic energy in joules of a 4 kg mass moving at 14 m/s.

Solution: KE = 1⁄2mv2 = 0.5·4·142 = 392.0.


Answer: 1152

Prompt: Compute the kinetic energy in joules of a 9 kg mass moving at 16 m/s.

Solution: KE = 1⁄2mv2 = 0.5·9·162 = 1152.0.


Answer: 294

Prompt: Compute the kinetic energy in joules of a 12 kg mass moving at 7 m/s.

Solution: KE = 1⁄2mv2 = 0.5·12·72 = 294.0.


Answer: 1014

Prompt: Compute the kinetic energy in joules of a 12 kg mass moving at 13 m/s.

Solution: KE = 1⁄2mv2 = 0.5·12·132 = 1014.0.


Answer: 1851.5

Prompt: Compute the kinetic energy in joules of a 7 kg mass moving at 23 m/s.

Solution: KE = 1⁄2mv2 = 0.5·7·232 = 1851.5.


Answer: 2592

Prompt: Compute the kinetic energy in joules of a 9 kg mass moving at 24 m/s.

Solution: KE = 1⁄2mv2 = 0.5·9·242 = 2592.0.


Answer: 661.5

Prompt: Compute the kinetic energy in joules of a 3 kg mass moving at 21 m/s.

Solution: KE = 1⁄2mv2 = 0.5·3·212 = 661.5.


Answer: 1083

Prompt: Compute the kinetic energy in joules of a 6 kg mass moving at 19 m/s.

Solution: KE = 1⁄2mv2 = 0.5·6·192 = 1083.0.


Answer: 196

Prompt: Compute the kinetic energy in joules of a 8 kg mass moving at 7 m/s.

Solution: KE = 1⁄2mv2 = 0.5·8·72 = 196.0.


Answer: 1470

Prompt: Compute the kinetic energy in joules of a 15 kg mass moving at 14 m/s.

Solution: KE = 1⁄2mv2 = 0.5·15·142 = 1470.0.


Answer: 2178

Prompt: Compute the kinetic energy in joules of a 9 kg mass moving at 22 m/s.

Solution: KE = 1⁄2mv2 = 0.5·9·222 = 2178.0.


Answer: 729

Prompt: Compute the kinetic energy in joules of a 18 kg mass moving at 9 m/s.

Solution: KE = 1⁄2mv2 = 0.5·18·92 = 729.0.


Answer: 4312

Prompt: Compute the kinetic energy in joules of a 11 kg mass moving at 28 m/s.

Solution: KE = 1⁄2mv2 = 0.5·11·282 = 4312.0.


Answer: 81

Prompt: Compute the kinetic energy in joules of a 2 kg mass moving at 9 m/s.

Solution: KE = 1⁄2mv2 = 0.5·2·92 = 81.0.


Answer: 729

Prompt: Compute the kinetic energy in joules of a 2 kg mass moving at 27 m/s.

Solution: KE = 1⁄2mv2 = 0.5·2·272 = 729.0.


Answer: 1078

Prompt: Compute the kinetic energy in joules of a 11 kg mass moving at 14 m/s.

Solution: KE = 1⁄2mv2 = 0.5·11·142 = 1078.0.


Answer: 72

Prompt: Compute the kinetic energy in joules of a 9 kg mass moving at 4 m/s.

Solution: KE = 1⁄2mv2 = 0.5·9·42 = 72.0.


Answer: 20

Prompt: A current of 1 A flows through 20 Ω. What is the voltage drop in volts?

Solution: V = IR = 1·20 = 20.


Answer: 16

Prompt: A current of 2 A flows through 8 Ω. What is the voltage drop in volts?

Solution: V = IR = 2·8 = 16.


Answer: 102

Prompt: A current of 6 A flows through 17 Ω. What is the voltage drop in volts?

Solution: V = IR = 6·17 = 102.


Answer: 55

Prompt: A current of 2.5 A flows through 22 Ω. What is the voltage drop in volts?

Solution: V = IR = 2.5·22 = 55.0.


Answer: 18

Prompt: A current of 3 A flows through 6 Ω. What is the voltage drop in volts?

Solution: V = IR = 3·6 = 18.


Answer: 9

Prompt: A current of 1.5 A flows through 6 Ω. What is the voltage drop in volts?

Solution: V = IR = 1.5·6 = 9.0.


Answer: 25

Prompt: A current of 1 A flows through 25 Ω. What is the voltage drop in volts?

Solution: V = IR = 1·25 = 25.


Answer: 76

Prompt: A current of 4 A flows through 19 Ω. What is the voltage drop in volts?

Solution: V = IR = 4·19 = 76.


Answer: 37.5

Prompt: A current of 1.5 A flows through 25 Ω. What is the voltage drop in volts?

Solution: V = IR = 1.5·25 = 37.5.


Answer: 75

Prompt: A current of 5 A flows through 15 Ω. What is the voltage drop in volts?

Solution: V = IR = 5·15 = 75.


Answer: 70

Prompt: A current of 5 A flows through 14 Ω. What is the voltage drop in volts?

Solution: V = IR = 5·14 = 70.


Answer: 18

Prompt: A current of 2 A flows through 9 Ω. What is the voltage drop in volts?

Solution: V = IR = 2·9 = 18.


Answer: 13.5

Prompt: A current of 0.75 A flows through 18 Ω. What is the voltage drop in volts?

Solution: V = IR = 0.75·18 = 13.5.


Answer: 12

Prompt: A current of 3 A flows through 4 Ω. What is the voltage drop in volts?

Solution: V = IR = 3·4 = 12.


Answer: 1.25

Prompt: A current of 0.25 A flows through 5 Ω. What is the voltage drop in volts?

Solution: V = IR = 0.25·5 = 1.25.


Answer: 16.5

Prompt: A current of 0.75 A flows through 22 Ω. What is the voltage drop in volts?

Solution: V = IR = 0.75·22 = 16.5.


Answer: 17.25

Prompt: A current of 0.75 A flows through 23 Ω. What is the voltage drop in volts?

Solution: V = IR = 0.75·23 = 17.25.


Answer: 52.5

Prompt: A current of 2.5 A flows through 21 Ω. What is the voltage drop in volts?

Solution: V = IR = 2.5·21 = 52.5.


Answer: 270

Prompt: A sample has density 2700 kg/m3 and volume 0.1 m3. What is its mass in kg?

Solution: m = ρV = 2700·0.1 = 270.0.


Answer: 1930

Prompt: A sample has density 19300 kg/m3 and volume 0.1 m3. What is its mass in kg?

Solution: m = ρV = 19300·0.1 = 1930.0.


Answer: 10

Prompt: A sample has density 1000 kg/m3 and volume 0.01 m3. What is its mass in kg?

Solution: m = ρV = 1000·0.01 = 10.0.


Answer: 135

Prompt: A sample has density 2700 kg/m3 and volume 0.05 m3. What is its mass in kg?

Solution: m = ρV = 2700·0.05 = 135.0.


Answer: 100

Prompt: A sample has density 1000 kg/m3 and volume 0.1 m3. What is its mass in kg?

Solution: m = ρV = 1000·0.1 = 100.0.


Answer: 780

Prompt: A sample has density 7800 kg/m3 and volume 0.1 m3. What is its mass in kg?

Solution: m = ρV = 7800·0.1 = 780.0.


Answer: 78

Prompt: A sample has density 7800 kg/m3 and volume 0.01 m3. What is its mass in kg?

Solution: m = ρV = 7800·0.01 = 78.0.


Answer: 20

Prompt: A sample has density 1000 kg/m3 and volume 0.02 m3. What is its mass in kg?

Solution: m = ρV = 1000·0.02 = 20.0.


Answer: 390

Prompt: A sample has density 7800 kg/m3 and volume 0.05 m3. What is its mass in kg?

Solution: m = ρV = 7800·0.05 = 390.0.


Answer: 27

Prompt: A sample has density 2700 kg/m3 and volume 0.01 m3. What is its mass in kg?

Solution: m = ρV = 2700·0.01 = 27.0.


Answer: 515468

Prompt: An ideal gas has n=2 mol, T=310 K, V=0.01 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 2.0·8.314·310/0.01 = 515468.0.


Answer: 49551.44

Prompt: An ideal gas has n=1 mol, T=298 K, V=0.05 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 1.0·8.314·298/0.05 = 49551.44.


Answer: 45394.44

Prompt: An ideal gas has n=1 mol, T=273 K, V=0.05 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 1.0·8.314·273/0.05 = 45394.44.


Answer: 90788.88

Prompt: An ideal gas has n=2 mol, T=273 K, V=0.05 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 2.0·8.314·273/0.05 = 90788.88.


Answer: 247757.2

Prompt: An ideal gas has n=2 mol, T=298 K, V=0.02 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 2.0·8.314·298/0.02 = 247757.2.


Answer: 145495

Prompt: An ideal gas has n=0.5 mol, T=350 K, V=0.01 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 0.5·8.314·350/0.01 = 145495.0.


Answer: 24775.72

Prompt: An ideal gas has n=0.5 mol, T=298 K, V=0.05 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 0.5·8.314·298/0.05 = 24775.72.


Answer: 116396

Prompt: An ideal gas has n=2 mol, T=350 K, V=0.05 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 2.0·8.314·350/0.05 = 116396.0.


Answer: 64433.5

Prompt: An ideal gas has n=0.5 mol, T=310 K, V=0.02 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 0.5·8.314·310/0.02 = 64433.5.


Answer: 247757.2

Prompt: An ideal gas has n=1 mol, T=298 K, V=0.01 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 1.0·8.314·298/0.01 = 247757.2.


Answer: 226972.2

Prompt: An ideal gas has n=1 mol, T=273 K, V=0.01 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 1.0·8.314·273/0.01 = 226972.2.


Answer: 113486.1

Prompt: An ideal gas has n=0.5 mol, T=273 K, V=0.01 m3. Using R=8.314 J/(mol·K), compute P in pascals.

Solution: P = nRT/V = 0.5·8.314·273/0.01 = 113486.1.


Answer: 17.87202381

m(H2)=2.0 g
stoichiometry 2:2 for H2:H2O

Convert to moles
Apply mole ratio
Convert to grams

Prompt: For 2 H2 + O2 → 2 H2O, how many grams of water form from 2 g of H2 if oxygen is in excess? Use M(H2)=2.016 g/mol and M(H2O)=18.015 g/mol.

Solution: mol H2 = 2.0/2.016 = 0.9920634920634921. mol H2O = mol H2 = 0.9920634920634921. mass = 0.9920634920634921·18.015 = 17.87202380952381.


Answer: 53.61607143

m(H2)=6.0 g
stoichiometry 2:2 for H2:H2O

Convert to moles
Apply mole ratio
Convert to grams

Prompt: For 2 H2 + O2 → 2 H2O, how many grams of water form from 6 g of H2 if oxygen is in excess? Use M(H2)=2.016 g/mol and M(H2O)=18.015 g/mol.

Solution: mol H2 = 6.0/2.016 = 2.9761904761904763. mol H2O = mol H2 = 2.9761904761904763. mass = 2.9761904761904763·18.015 = 53.61607142857143.


Answer: 35.74404762

m(H2)=4.0 g
stoichiometry 2:2 for H2:H2O

Convert to moles
Apply mole ratio
Convert to grams

Prompt: For 2 H2 + O2 → 2 H2O, how many grams of water form from 4 g of H2 if oxygen is in excess? Use M(H2)=2.016 g/mol and M(H2O)=18.015 g/mol.

Solution: mol H2 = 4.0/2.016 = 1.9841269841269842. mol H2O = mol H2 = 1.9841269841269842. mass = 1.9841269841269842·18.015 = 35.74404761904762.


Answer: 71.48809524

m(H2)=8.0 g
stoichiometry 2:2 for H2:H2O

Convert to moles
Apply mole ratio
Convert to grams

Prompt: For 2 H2 + O2 → 2 H2O, how many grams of water form from 8 g of H2 if oxygen is in excess? Use M(H2)=2.016 g/mol and M(H2O)=18.015 g/mol.

Solution: mol H2 = 8.0/2.016 = 3.9682539682539684. mol H2O = mol H2 = 3.9682539682539684. mass = 3.9682539682539684·18.015 = 71.48809523809524.


Answer: 4

25 °C
ideal dilute solution
complete dissociation

Prompt: A strong acid is fully dissociated at [0.0001 ] mol/L H+. What is the pH?

Solution: pH = -log10[H+] = -log10(0.0001) = 4.0.


Answer: 2

25 °C
ideal dilute solution
complete dissociation

Prompt: A strong acid is fully dissociated at [0.01 ] mol/L H+. What is the pH?

Solution: pH = -log10[H+] = -log10(0.01) = 2.0.


Answer: 1

25 °C
ideal dilute solution
complete dissociation

Prompt: A strong acid is fully dissociated at [0.1 ] mol/L H+. What is the pH?

Solution: pH = -log10[H+] = -log10(0.1) = 1.0.


Answer: 3

25 °C
ideal dilute solution
complete dissociation

Prompt: A strong acid is fully dissociated at [0.001 ] mol/L H+. What is the pH?

Solution: pH = -log10[H+] = -log10(0.001) = 3.0.


Answer: 3870

Prompt: Heat 2 kg of a material with c=387 J/(kg·K) through ΔT=5 K. What energy in joules is required?

Solution: Q = mcΔT = 2.0·387·5 = 3870.0.


Answer: 16736

Prompt: Heat 0.2 kg of a material with c=4184 J/(kg·K) through ΔT=20 K. What energy in joules is required?

Solution: Q = mcΔT = 0.2·4184·20 = 16736.0.


Answer: 41840

Prompt: Heat 1 kg of a material with c=4184 J/(kg·K) through ΔT=10 K. What energy in joules is required?

Solution: Q = mcΔT = 1.0·4184·10 = 41840.0.


Answer: 52300

Prompt: Heat 0.5 kg of a material with c=4184 J/(kg·K) through ΔT=25 K. What energy in joules is required?

Solution: Q = mcΔT = 0.5·4184·25 = 52300.0.


Answer: 167360

Prompt: Heat 2 kg of a material with c=4184 J/(kg·K) through ΔT=20 K. What energy in joules is required?

Solution: Q = mcΔT = 2.0·4184·20 = 167360.0.


Answer: 9000

Prompt: Heat 0.5 kg of a material with c=900 J/(kg·K) through ΔT=20 K. What energy in joules is required?

Solution: Q = mcΔT = 0.5·900·20 = 9000.0.


Answer: 41840

Prompt: Heat 0.5 kg of a material with c=4184 J/(kg·K) through ΔT=20 K. What energy in joules is required?

Solution: Q = mcΔT = 0.5·4184·20 = 41840.0.


Answer: 20920

Prompt: Heat 0.2 kg of a material with c=4184 J/(kg·K) through ΔT=25 K. What energy in joules is required?

Solution: Q = mcΔT = 0.2·4184·25 = 20920.0.


Answer: 8368

Prompt: Heat 0.2 kg of a material with c=4184 J/(kg·K) through ΔT=10 K. What energy in joules is required?

Solution: Q = mcΔT = 0.2·4184·10 = 8368.0.


Answer: 2700

Prompt: Heat 0.2 kg of a material with c=900 J/(kg·K) through ΔT=15 K. What energy in joules is required?

Solution: Q = mcΔT = 0.2·900·15 = 2700.0.


Answer: 36000

Prompt: Heat 2 kg of a material with c=900 J/(kg·K) through ΔT=20 K. What energy in joules is required?

Solution: Q = mcΔT = 2.0·900·20 = 36000.0.


Answer: 83680

Prompt: Heat 2 kg of a material with c=4184 J/(kg·K) through ΔT=10 K. What energy in joules is required?

Solution: Q = mcΔT = 2.0·4184·10 = 83680.0.


Answer: 20920

Prompt: Heat 1 kg of a material with c=4184 J/(kg·K) through ΔT=5 K. What energy in joules is required?

Solution: Q = mcΔT = 1.0·4184·5 = 20920.0.


Answer: 0.125

Prompt: A wave travels at 3e+08 m/s with frequency 2.4e+09 Hz. What is its wavelength in metres?

Solution: λ = v/f = 300000000.0/2400000000.0 = 0.125.


Answer: 1.416666667e-07

Prompt: A wave travels at 340 m/s with frequency 2.4e+09 Hz. What is its wavelength in metres?

Solution: λ = v/f = 340/2400000000.0 = 1.4166666666666668e-07.


Answer: 0.6640625

Prompt: A wave travels at 340 m/s with frequency 512 Hz. What is its wavelength in metres?

Solution: λ = v/f = 340/512 = 0.6640625.


Answer: 1171875

Prompt: A wave travels at 3e+08 m/s with frequency 256 Hz. What is its wavelength in metres?

Solution: λ = v/f = 300000000.0/256 = 1171875.0.


Answer: 300000

Prompt: A wave travels at 3e+08 m/s with frequency 1000 Hz. What is its wavelength in metres?

Solution: λ = v/f = 300000000.0/1000 = 300000.0.


Answer: 1.328125

Prompt: A wave travels at 340 m/s with frequency 256 Hz. What is its wavelength in metres?

Solution: λ = v/f = 340/256 = 1.328125.


Answer: 3.409090909

Prompt: A wave travels at 1500 m/s with frequency 440 Hz. What is its wavelength in metres?

Solution: λ = v/f = 1500/440 = 3.409090909090909.


Answer: 681818.1818

Prompt: A wave travels at 3e+08 m/s with frequency 440 Hz. What is its wavelength in metres?

Solution: λ = v/f = 300000000.0/440 = 681818.1818181818.


Answer: 0.75

Mendelian independent segregation
Complete dominance

Write the Punnett square
Count the requested class

Prompt: In a monohybrid Aa × Aa cross with complete dominance, what fraction of offspring are expected to show the dominant phenotype?

Solution: Genotypes: 1 AA : 2 Aa : 1 aa. Dominant phenotype = 3/4.


Answer: 0.5

Mendelian independent segregation
Complete dominance

Write the Punnett square
Count the requested class

Prompt: In a monohybrid Aa × Aa cross, what fraction of offspring are heterozygous?

Solution: Punnett square: AA, Aa, aA, aa. Heterozygotes = 2/4 = 1/2.


Answer: 200

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 1600 nuclei has half-life 5 years. How many remain after 15 years?

Solution: N = N0 · (1/2)^(t/T) = 1600 · 0.5^(15/5) = 200.0.


Answer: 500

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 1000 nuclei has half-life 4 years. How many remain after 4 years?

Solution: N = N0 · (1/2)^(t/T) = 1000 · 0.5^(4/4) = 500.0.


Answer: 400

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 1600 nuclei has half-life 5 years. How many remain after 10 years?

Solution: N = N0 · (1/2)^(t/T) = 1600 · 0.5^(10/5) = 400.0.


Answer: 200

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 800 nuclei has half-life 4 years. How many remain after 8 years?

Solution: N = N0 · (1/2)^(t/T) = 800 · 0.5^(8/4) = 200.0.


Answer: 100

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 800 nuclei has half-life 5 years. How many remain after 15 years?

Solution: N = N0 · (1/2)^(t/T) = 800 · 0.5^(15/5) = 100.0.


Answer: 250

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 1000 nuclei has half-life 4 years. How many remain after 8 years?

Solution: N = N0 · (1/2)^(t/T) = 1000 · 0.5^(8/4) = 250.0.


Answer: 200

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 800 nuclei has half-life 2 years. How many remain after 4 years?

Solution: N = N0 · (1/2)^(t/T) = 800 · 0.5^(4/2) = 200.0.


Answer: 800

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 1600 nuclei has half-life 5 years. How many remain after 5 years?

Solution: N = N0 · (1/2)^(t/T) = 1600 · 0.5^(5/5) = 800.0.


Answer: 400

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 800 nuclei has half-life 5 years. How many remain after 5 years?

Solution: N = N0 · (1/2)^(t/T) = 800 · 0.5^(5/5) = 400.0.


Answer: 800

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 1600 nuclei has half-life 4 years. How many remain after 4 years?

Solution: N = N0 · (1/2)^(t/T) = 1600 · 0.5^(4/4) = 800.0.


Answer: 500

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 1000 nuclei has half-life 2 years. How many remain after 2 years?

Solution: N = N0 · (1/2)^(t/T) = 1000 · 0.5^(2/2) = 500.0.


Answer: 125

Compute the number of half-lives
Multiply by 1/2 each time

Prompt: A sample of 1000 nuclei has half-life 5 years. How many remain after 15 years?

Solution: N = N0 · (1/2)^(t/T) = 1000 · 0.5^(15/5) = 125.0.


Answer: 0.2407500841

Sun-dominated two-body problem
T in years, a in AU

Prompt: Using Kepler's third law in AU and years (T2 = a3), estimate the orbital period of a planet with semi-major axis 0.387 AU.

Solution: T = a^(3/2) = 0.387^1.5 = 0.24075008411213486 years.


Answer: 1.881384018

Sun-dominated two-body problem
T in years, a in AU

Prompt: Using Kepler's third law in AU and years (T2 = a3), estimate the orbital period of a planet with semi-major axis 1.524 AU.

Solution: T = a^(3/2) = 1.524^1.5 = 1.8813840182163768 years.


Answer: 0.6147626103

Sun-dominated two-body problem
T in years, a in AU

Prompt: Using Kepler's third law in AU and years (T2 = a3), estimate the orbital period of a planet with semi-major axis 0.723 AU.

Solution: T = a^(3/2) = 0.723^1.5 = 0.6147626102813996 years.


Answer: 0.4

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.05 L of a 2 M stock to 0.25 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 2.0·0.05/0.25 = 0.4.


Answer: 0.2

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.05 L of a 2 M stock to 0.5 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 2.0·0.05/0.5 = 0.2.


Answer: 1.2

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.05 L of a 6 M stock to 0.25 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 6.0·0.05/0.25 = 1.2000000000000002.


Answer: 0.6

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.01 L of a 6 M stock to 0.1 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 6.0·0.01/0.1 = 0.6.


Answer: 3

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.025 L of a 12 M stock to 0.1 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 12.0·0.025/0.1 = 3.0000000000000004.


Answer: 0.25

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.025 L of a 1 M stock to 0.1 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 1.0·0.025/0.1 = 0.25.


Answer: 0.12

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.01 L of a 6 M stock to 0.5 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 6.0·0.01/0.5 = 0.12.


Answer: 1.2

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.05 L of a 12 M stock to 0.5 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 12.0·0.05/0.5 = 1.2000000000000002.


Answer: 6

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.05 L of a 12 M stock to 0.1 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 12.0·0.05/0.1 = 6.000000000000001.


Answer: 0.02

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.01 L of a 1 M stock to 0.5 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 1.0·0.01/0.5 = 0.02.


Answer: 1

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.05 L of a 2 M stock to 0.1 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 2.0·0.05/0.1 = 1.0.


Answer: 0.6

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.05 L of a 6 M stock to 0.5 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 6.0·0.05/0.5 = 0.6000000000000001.


Answer: 0.5

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.025 L of a 2 M stock to 0.1 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 2.0·0.025/0.1 = 0.5.


Answer: 0.1

Conservative solute
Volumes additive as stated

Prompt: Dilute 0.025 L of a 1 M stock to 0.25 L. What is the final molarity?

Solution: M1V1 = M2V2 ⇒ M2 = 1.0·0.025/0.25 = 0.1.


Answer: 64

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: An algorithm performs n log2(n) comparisons. For n=16, how many comparisons?

Solution: n log2 n = 16·4.0 = 64.0.


Answer: 2080

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: A nested loop runs i=1..n, j=1..i. How many (i,j) pairs for n=64?

Solution: Sum_{i=1..n} i = n(n+1)/2 = 2080.0.


Answer: 5

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: How many bits are required to represent the integer 32 in binary (floor(log2 n))?

Solution: floor(log2(32)) = 5.


Answer: 13

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: Convert the binary integer 1101 to decimal.

Solution: 11012 = 1310.


Answer: 2048

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: An algorithm performs n log2(n) comparisons. For n=256, how many comparisons?

Solution: n log2 n = 256·8.0 = 2048.0.


Answer: 8256

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: A nested loop runs i=1..n, j=1..i. How many (i,j) pairs for n=128?

Solution: Sum_{i=1..n} i = n(n+1)/2 = 8256.0.


Answer: 8

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: How many bits are required to represent the integer 256 in binary (floor(log2 n))?

Solution: floor(log2(256)) = 8.


Answer: 160

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: An algorithm performs n log2(n) comparisons. For n=32, how many comparisons?

Solution: n log2 n = 32·5.0 = 160.0.


Answer: 4

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: How many bits are required to represent the integer 16 in binary (floor(log2 n))?

Solution: floor(log2(16)) = 4.


Answer: 58

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: Convert the binary integer 111010 to decimal.

Solution: 1110102 = 5810.


Answer: 896

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: An algorithm performs n log2(n) comparisons. For n=128, how many comparisons?

Solution: n log2 n = 128·7.0 = 896.0.


Answer: 528

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: A nested loop runs i=1..n, j=1..i. How many (i,j) pairs for n=32?

Solution: Sum_{i=1..n} i = n(n+1)/2 = 528.0.


Answer: 6

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: How many bits are required to represent the integer 64 in binary (floor(log2 n))?

Solution: floor(log2(64)) = 6.


Answer: 14

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: Convert the binary integer 1110 to decimal.

Solution: 11102 = 1410.


Answer: 52

Exact integer or binary logarithm as specified

Apply the stated formula or place-value conversion

Prompt: Convert the binary integer 110100 to decimal.

Solution: 1101002 = 5210.


Answer: 5000000

Prompt: A tensile force of 5000 N acts on a cross-section of 0.001 m2. What is the axial stress in pascals?

Solution: σ = F/A = 5000/0.001 = 5000000.0.


Answer: 2500000

Prompt: A tensile force of 2500 N acts on a cross-section of 0.001 m2. What is the axial stress in pascals?

Solution: σ = F/A = 2500/0.001 = 2500000.0.


Answer: 1250000

Prompt: A tensile force of 2500 N acts on a cross-section of 0.002 m2. What is the axial stress in pascals?

Solution: σ = F/A = 2500/0.002 = 1250000.0.


Answer: 2500000

Prompt: A tensile force of 5000 N acts on a cross-section of 0.002 m2. What is the axial stress in pascals?

Solution: σ = F/A = 5000/0.002 = 2500000.0.


Answer: 500000

Prompt: A tensile force of 1000 N acts on a cross-section of 0.002 m2. What is the axial stress in pascals?

Solution: σ = F/A = 1000/0.002 = 500000.0.


Answer: 1000000

Prompt: A tensile force of 5000 N acts on a cross-section of 0.005 m2. What is the axial stress in pascals?

Solution: σ = F/A = 5000/0.005 = 1000000.0.


Answer: 2000000

Prompt: A tensile force of 10000 N acts on a cross-section of 0.005 m2. What is the axial stress in pascals?

Solution: σ = F/A = 10000/0.005 = 2000000.0.


Answer: 500000

Prompt: A tensile force of 2500 N acts on a cross-section of 0.005 m2. What is the axial stress in pascals?

Solution: σ = F/A = 2500/0.005 = 500000.0.


Answer: 200000

Prompt: A tensile force of 1000 N acts on a cross-section of 0.005 m2. What is the axial stress in pascals?

Solution: σ = F/A = 1000/0.005 = 200000.0.


Answer: 5000000

Prompt: A tensile force of 10000 N acts on a cross-section of 0.002 m2. What is the axial stress in pascals?

Solution: σ = F/A = 10000/0.002 = 5000000.0.


Answer: 30

One unit per replicate
No blocking factor in this count

Prompt: You will test 6 fertilizer treatments with 5 independent replicates each, fully randomized. How many experimental units do you need?

Solution: units = treatments × replicates = 6×5 = 30.


Answer: 16

One unit per replicate
No blocking factor in this count

Prompt: You will test 4 fertilizer treatments with 4 independent replicates each, fully randomized. How many experimental units do you need?

Solution: units = treatments × replicates = 4×4 = 16.


Answer: 18

One unit per replicate
No blocking factor in this count

Prompt: You will test 6 fertilizer treatments with 3 independent replicates each, fully randomized. How many experimental units do you need?

Solution: units = treatments × replicates = 6×3 = 18.


Answer: 15

One unit per replicate
No blocking factor in this count

Prompt: You will test 5 fertilizer treatments with 3 independent replicates each, fully randomized. How many experimental units do you need?

Solution: units = treatments × replicates = 5×3 = 15.


Answer: 24

One unit per replicate
No blocking factor in this count

Prompt: You will test 6 fertilizer treatments with 4 independent replicates each, fully randomized. How many experimental units do you need?

Solution: units = treatments × replicates = 6×4 = 24.


Answer: 20

One unit per replicate
No blocking factor in this count

Prompt: You will test 4 fertilizer treatments with 5 independent replicates each, fully randomized. How many experimental units do you need?

Solution: units = treatments × replicates = 4×5 = 20.


Answer: 1870.65

Prompt: An ideal gas has n=3 mol, T=300 K, V=4 m^3. Using R=8.314 J/(mol·K), what is P in pascals from PV=nRT?

Solution: P = nRT/V = 3*8.314*300/4 = 1870.65.


Answer: 453.9444

Prompt: An ideal gas has n=2 mol, T=273 K, V=10 m^3. Using R=8.314 J/(mol·K), what is P in pascals from PV=nRT?

Solution: P = nRT/V = 2*8.314*273/10 = 453.94440000000003.


Answer: 415.7

Prompt: An ideal gas has n=1 mol, T=250 K, V=5 m^3. Using R=8.314 J/(mol·K), what is P in pascals from PV=nRT?

Solution: P = nRT/V = 1*8.314*250/5 = 415.7.


Answer: 3741.3

Prompt: An ideal gas has n=3 mol, T=300 K, V=2 m^3. Using R=8.314 J/(mol·K), what is P in pascals from PV=nRT?

Solution: P = nRT/V = 3*8.314*300/2 = 3741.3.


Answer: 2182.425

Prompt: An ideal gas has n=3 mol, T=350 K, V=4 m^3. Using R=8.314 J/(mol·K), what is P in pascals from PV=nRT?

Solution: P = nRT/V = 3*8.314*350/4 = 2182.425.


Answer: 5819.8

Prompt: An ideal gas has n=2 mol, T=350 K, V=1 m^3. Using R=8.314 J/(mol·K), what is P in pascals from PV=nRT?

Solution: P = nRT/V = 2*8.314*350/1 = 5819.8.


Answer: 2269.722

Prompt: An ideal gas has n=2 mol, T=273 K, V=2 m^3. Using R=8.314 J/(mol·K), what is P in pascals from PV=nRT?

Solution: P = nRT/V = 2*8.314*273/2 = 2269.722.


Answer: 581.98

Prompt: An ideal gas has n=2 mol, T=350 K, V=10 m^3. Using R=8.314 J/(mol·K), what is P in pascals from PV=nRT?

Solution: P = nRT/V = 2*8.314*350/10 = 581.98.


Answer: 1361.8332

Prompt: An ideal gas has n=3 mol, T=273 K, V=5 m^3. Using R=8.314 J/(mol·K), what is P in pascals from PV=nRT?

Solution: P = nRT/V = 3*8.314*273/5 = 1361.8332.


Answer: 2494.2

Prompt: An ideal gas has n=2 mol, T=300 K, V=2 m^3. Using R=8.314 J/(mol·K), what is P in pascals from PV=nRT?

Solution: P = nRT/V = 2*8.314*300/2 = 2494.2.


Answer: 4.5

Prompt: A sample has mass 18 kg and volume 4 m^3. What is its density in kg/m^3?

Solution: ρ = m/V = 18/4 = 4.5.


Answer: 14

Prompt: A sample has mass 7 kg and volume 0.5 m^3. What is its density in kg/m^3?

Solution: ρ = m/V = 7/0.5 = 14.0.


Answer: 2.2

Prompt: A sample has mass 11 kg and volume 5 m^3. What is its density in kg/m^3?

Solution: ρ = m/V = 11/5 = 2.2.


Answer: 24

Prompt: A sample has mass 12 kg and volume 0.5 m^3. What is its density in kg/m^3?

Solution: ρ = m/V = 12/0.5 = 24.0.


Answer: 4.5

Prompt: A sample has mass 9 kg and volume 2 m^3. What is its density in kg/m^3?

Solution: ρ = m/V = 9/2 = 4.5.


Answer: 9

Prompt: A sample has mass 9 kg and volume 1 m^3. What is its density in kg/m^3?

Solution: ρ = m/V = 9/1 = 9.0.


Answer: 12

Prompt: A sample has mass 6 kg and volume 0.5 m^3. What is its density in kg/m^3?

Solution: ρ = m/V = 6/0.5 = 12.0.


Answer: 3

Prompt: A sample has mass 3 kg and volume 1 m^3. What is its density in kg/m^3?

Solution: ρ = m/V = 3/1 = 3.0.


Answer: 3.4

Prompt: A sample has mass 17 kg and volume 5 m^3. What is its density in kg/m^3?

Solution: ρ = m/V = 17/5 = 3.4.


Answer: 1

Prompt: A sample has mass 4 kg and volume 4 m^3. What is its density in kg/m^3?

Solution: ρ = m/V = 4/4 = 1.0.


Answer: 17.14285714

Thin-lens equation
Positive real-object distances

Prompt: A thin converging lens has f=12 cm. A real object is u=40 cm away. Using 1/f = 1/v + 1/u with positive distances, what is v in cm?

Solution: 1/v = 1/f - 1/u = 1/12 - 1/40 so v = 17.142857142857146.


Answer: 10.90909091

Thin-lens equation
Positive real-object distances

Prompt: A thin converging lens has f=8 cm. A real object is u=30 cm away. Using 1/f = 1/v + 1/u with positive distances, what is v in cm?

Solution: 1/v = 1/f - 1/u = 1/8 - 1/30 so v = 10.909090909090908.


Answer: 20

Thin-lens equation
Positive real-object distances

Prompt: A thin converging lens has f=12 cm. A real object is u=30 cm away. Using 1/f = 1/v + 1/u with positive distances, what is v in cm?

Solution: 1/v = 1/f - 1/u = 1/12 - 1/30 so v = 20.0.


Answer: 13.33333333

Thin-lens equation
Positive real-object distances

Prompt: A thin converging lens has f=8 cm. A real object is u=20 cm away. Using 1/f = 1/v + 1/u with positive distances, what is v in cm?

Solution: 1/v = 1/f - 1/u = 1/8 - 1/20 so v = 13.333333333333334.


Answer: 7.5

Thin-lens equation
Positive real-object distances

Prompt: A thin converging lens has f=5 cm. A real object is u=15 cm away. Using 1/f = 1/v + 1/u with positive distances, what is v in cm?

Solution: 1/v = 1/f - 1/u = 1/5 - 1/15 so v = 7.499999999999998.


Answer: 30

Thin-lens equation
Positive real-object distances

Prompt: A thin converging lens has f=12 cm. A real object is u=20 cm away. Using 1/f = 1/v + 1/u with positive distances, what is v in cm?

Solution: 1/v = 1/f - 1/u = 1/12 - 1/20 so v = 30.000000000000007.


Answer: 10

Thin-lens equation
Positive real-object distances

Prompt: A thin converging lens has f=8 cm. A real object is u=40 cm away. Using 1/f = 1/v + 1/u with positive distances, what is v in cm?

Solution: 1/v = 1/f - 1/u = 1/8 - 1/40 so v = 10.0.


Answer: 60

Thin-lens equation
Positive real-object distances

Prompt: A thin converging lens has f=12 cm. A real object is u=15 cm away. Using 1/f = 1/v + 1/u with positive distances, what is v in cm?

Solution: 1/v = 1/f - 1/u = 1/12 - 1/15 so v = 60.000000000000014.


Answer: 3.973835306

Prompt: A simple pendulum has length 4 m. Take g=10 m/s2 and T=2π√(L/g). What is the period in seconds?

Solution: T=2π√(4/10) = 3.9738353063184406.


Answer: 1.884955592

Prompt: A simple pendulum has length 0.9 m. Take g=10 m/s2 and T=2π√(L/g). What is the period in seconds?

Solution: T=2π√(0.9/10) = 1.8849555921538759.


Answer: 1.256637061

Prompt: A simple pendulum has length 0.4 m. Take g=10 m/s2 and T=2π√(L/g). What is the period in seconds?

Solution: T=2π√(0.4/10) = 1.2566370614359172.


Answer: 3.141592654

Prompt: A simple pendulum has length 2.5 m. Take g=10 m/s2 and T=2π√(L/g). What is the period in seconds?

Solution: T=2π√(2.5/10) = 3.141592653589793.


Answer: 4.002e-05

Prompt: Two point masses 500000 kg and 300000 kg are 500 m apart. Using G=6.67e-11, what is the Newtonian force magnitude in newtons?

Solution: F=G m1 m2 / r^2 = 4.001999999999999e-05.


Answer: 0.0002001

Prompt: Two point masses 800000 kg and 600000 kg are 400 m apart. Using G=6.67e-11, what is the Newtonian force magnitude in newtons?

Solution: F=G m1 m2 / r^2 = 0.00020009999999999998.


Answer: 0.00010005

Prompt: Two point masses 400000 kg and 600000 kg are 400 m apart. Using G=6.67e-11, what is the Newtonian force magnitude in newtons?

Solution: F=G m1 m2 / r^2 = 0.00010004999999999999.


Answer: 0.0004002

Prompt: Two point masses 800000 kg and 300000 kg are 200 m apart. Using G=6.67e-11, what is the Newtonian force magnitude in newtons?

Solution: F=G m1 m2 / r^2 = 0.00040019999999999997.


Answer: 0.0001250625

Prompt: Two point masses 500000 kg and 600000 kg are 400 m apart. Using G=6.67e-11, what is the Newtonian force magnitude in newtons?

Solution: F=G m1 m2 / r^2 = 0.00012506249999999998.


Answer: 5.0025e-05

Prompt: Two point masses 400000 kg and 300000 kg are 400 m apart. Using G=6.67e-11, what is the Newtonian force magnitude in newtons?

Solution: F=G m1 m2 / r^2 = 5.0024999999999996e-05.


Answer: 5.0025e-05

Prompt: Two point masses 200000 kg and 600000 kg are 400 m apart. Using G=6.67e-11, what is the Newtonian force magnitude in newtons?


Answer: 4

Prompt: 1 mol of solute is dissolved to 0.25 L of solution. What is the molarity in mol/L?

Solution: M = n/V = 1/0.25 = 4.0.


Answer: 1

Prompt: 0.5 mol of solute is dissolved to 0.5 L of solution. What is the molarity in mol/L?

Solution: M = n/V = 0.5/0.5 = 1.0.


Answer: 0.25

Prompt: 0.5 mol of solute is dissolved to 2 L of solution. What is the molarity in mol/L?

Solution: M = n/V = 0.5/2 = 0.25.


Answer: 0.125

Prompt: 0.25 mol of solute is dissolved to 2 L of solution. What is the molarity in mol/L?

Solution: M = n/V = 0.25/2 = 0.125.


Answer: 2

Prompt: 2 mol of solute is dissolved to 1 L of solution. What is the molarity in mol/L?

Solution: M = n/V = 2/1 = 2.0.


Answer: 1

Prompt: 2 mol of solute is dissolved to 2 L of solution. What is the molarity in mol/L?

Solution: M = n/V = 2/2 = 1.0.


Answer: 0.5

Prompt: 0.25 mol of solute is dissolved to 0.5 L of solution. What is the molarity in mol/L?

Solution: M = n/V = 0.25/0.5 = 0.5.


Answer: 0.415625

Prompt: Snell's law: n1 sin θ1 = n2 sin θ2. n1=1.33, n2=1.6, sin θ1=0.5. What is sin θ2?

Solution: sin θ2 = n1 sin θ1 / n2 = 0.415625.


Answer: 0.25

Prompt: Snell's law: n1 sin θ1 = n2 sin θ2. n1=1, n2=1.6, sin θ1=0.4. What is sin θ2?

Solution: sin θ2 = n1 sin θ1 / n2 = 0.25.


Answer: 0.1875

Prompt: Snell's law: n1 sin θ1 = n2 sin θ2. n1=1, n2=1.6, sin θ1=0.3. What is sin θ2?

Solution: sin θ2 = n1 sin θ1 / n2 = 0.18749999999999997.


Answer: 0.3325

Prompt: Snell's law: n1 sin θ1 = n2 sin θ2. n1=1.33, n2=1.6, sin θ1=0.4. What is sin θ2?

Solution: sin θ2 = n1 sin θ1 / n2 = 0.3325.


Answer: 5

Prompt: A sample starts at 80 units. After 4 half-lives, how many units remain (exponential decay, integer-power of two)?

Solution: N = N0 / 2^h = 80/2^4 = 5.0.


Answer: 10

Prompt: A sample starts at 160 units. After 4 half-lives, how many units remain (exponential decay, integer-power of two)?

Solution: N = N0 / 2^h = 160/2^4 = 10.0.


Answer: 80

Prompt: A sample starts at 320 units. After 2 half-lives, how many units remain (exponential decay, integer-power of two)?

Solution: N = N0 / 2^h = 320/2^2 = 80.0.


Answer: 40

Prompt: A sample starts at 160 units. After 2 half-lives, how many units remain (exponential decay, integer-power of two)?

Solution: N = N0 / 2^h = 160/2^2 = 40.0.


Answer: 160

Prompt: A sample starts at 320 units. After 1 half-lives, how many units remain (exponential decay, integer-power of two)?

Solution: N = N0 / 2^h = 320/2^1 = 160.0.


Answer: 80

Prompt: A sample starts at 160 units. After 1 half-lives, how many units remain (exponential decay, integer-power of two)?

Solution: N = N0 / 2^h = 160/2^1 = 80.0.


Answer: 3/4

Mendelian, independent, complete dominance

Prompt: Aa × Aa, complete dominance. What is P(dominant phenotype) as a reduced fraction a/b? Report a/b.

Solution: 3/4 of the Punnett cells match.


Answer: 1/2

Mendelian, independent, complete dominance

Prompt: Aa × aa testcross, complete dominance. What is P(recessive phenotype) as a reduced fraction a/b? Report a/b.

Solution: 1/2 of the Punnett cells match.


Answer: 9000

Prompt: Q=mcΔT. Mass 1 kg, c=900 J/(kg·K), ΔT=10 K. What is Q in joules?


Answer: 4500

Prompt: Q=mcΔT. Mass 1 kg, c=900 J/(kg·K), ΔT=5 K. What is Q in joules?


Answer: 83600

Prompt: Q=mcΔT. Mass 1 kg, c=4180 J/(kg·K), ΔT=20 K. What is Q in joules?


Answer: 83600

Prompt: Q=mcΔT. Mass 2 kg, c=4180 J/(kg·K), ΔT=10 K. What is Q in joules?


Answer: 967.5

Prompt: Q=mcΔT. Mass 0.5 kg, c=387 J/(kg·K), ΔT=5 K. What is Q in joules?


Answer: 41800

Prompt: Q=mcΔT. Mass 2 kg, c=4180 J/(kg·K), ΔT=5 K. What is Q in joules?


Answer: 10450

Prompt: Q=mcΔT. Mass 0.5 kg, c=4180 J/(kg·K), ΔT=5 K. What is Q in joules?


Answer: 20

Prompt: Archimedes: a fully submerged volume 0.002 m3 in freshwater (ρ=1000 kg/m3), g=10 m/s2. What is the buoyant force in newtons?


Answer: 50

Prompt: Archimedes: a fully submerged volume 0.005 m3 in freshwater (ρ=1000 kg/m3), g=10 m/s2. What is the buoyant force in newtons?


Answer: 100

Prompt: Archimedes: a fully submerged volume 0.01 m3 in freshwater (ρ=1000 kg/m3), g=10 m/s2. What is the buoyant force in newtons?


Answer: 3.9756e-19

Prompt: A photon has frequency 6e+14 Hz. Using E=hf and h=6.626e-34 J·s, what is E in joules?


Answer: 6.626e-19

Prompt: A photon has frequency 1e+15 Hz. Using E=hf and h=6.626e-34 J·s, what is E in joules?

Solution: E=hf=6.626e-34·1e+15=6.625999999999999e-19.


Answer: 3.313e-19

Prompt: A photon has frequency 5e+14 Hz. Using E=hf and h=6.626e-34 J·s, what is E in joules?


Answer: 0.2

Conservative solute
Volumes additive as given

Prompt: Dilute 0.1 L of 2 mol/L solution to 1 L. What is the new concentration in mol/L (C1V1=C2V2)?


Answer: 1

Conservative solute
Volumes additive as given

Prompt: Dilute 0.25 L of 2 mol/L solution to 0.5 L. What is the new concentration in mol/L (C1V1=C2V2)?


Answer: 0.75

Conservative solute
Volumes additive as given

Prompt: Dilute 0.25 L of 6 mol/L solution to 2 L. What is the new concentration in mol/L (C1V1=C2V2)?


Answer: 1

Conservative solute
Volumes additive as given

Prompt: Dilute 0.2 L of 10 mol/L solution to 2 L. What is the new concentration in mol/L (C1V1=C2V2)?


Answer: 0.2

Conservative solute
Volumes additive as given

Prompt: Dilute 0.1 L of 4 mol/L solution to 2 L. What is the new concentration in mol/L (C1V1=C2V2)?


Answer: 53

Linear calibration
No noise term in this item

Prompt: A calibrated sensor is modeled as y=7x+4. If the reading x=7, what is the modeled y?


Answer: 39

Linear calibration
No noise term in this item

Prompt: A calibrated sensor is modeled as y=7x+-3. If the reading x=6, what is the modeled y?


Answer: 18

Linear calibration
No noise term in this item

Prompt: A calibrated sensor is modeled as y=2x+2. If the reading x=8, what is the modeled y?


Answer: 20

Linear calibration
No noise term in this item

Prompt: A calibrated sensor is modeled as y=3x+2. If the reading x=6, what is the modeled y?


Answer: 61

Linear calibration
No noise term in this item

Prompt: A calibrated sensor is modeled as y=9x+-2. If the reading x=7, what is the modeled y?


Answer: 24

Linear calibration
No noise term in this item

Prompt: A calibrated sensor is modeled as y=4x+4. If the reading x=5, what is the modeled y?


Answer: 2.4

Prompt: On a distance-time graph, distance rises 12 m over 5 s of straight-line motion. What is the constant speed in m/s?

Solution: speed = Δs/Δt = 12/5 = 2.4.


Answer: 3

Prompt: On a distance-time graph, distance rises 12 m over 4 s of straight-line motion. What is the constant speed in m/s?

Solution: speed = Δs/Δt = 12/4 = 3.0.


Answer: 2

Prompt: On a distance-time graph, distance rises 10 m over 5 s of straight-line motion. What is the constant speed in m/s?

Solution: speed = Δs/Δt = 10/5 = 2.0.


Answer: 2

Prompt: On a distance-time graph, distance rises 8 m over 4 s of straight-line motion. What is the constant speed in m/s?

Solution: speed = Δs/Δt = 8/4 = 2.0.


Answer: 5

Prompt: On a distance-time graph, distance rises 10 m over 2 s of straight-line motion. What is the constant speed in m/s?

Solution: speed = Δs/Δt = 10/2 = 5.0.


Answer: 193.6

Prompt: A mass of 8 kg moves in a horizontal circle of radius 5 m at 11 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 8·112/5 = 193.6.


Answer: 343

Prompt: A mass of 7 kg moves in a horizontal circle of radius 1 m at 7 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 7·72/1 = 343.0.


Answer: 196

Prompt: A mass of 8 kg moves in a horizontal circle of radius 2 m at 7 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 8·72/2 = 196.0.


Answer: 66.66666667

Prompt: A mass of 8 kg moves in a horizontal circle of radius 3 m at 5 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 8·52/3 = 66.66666666666667.


Answer: 81.66666667

Prompt: A mass of 5 kg moves in a horizontal circle of radius 3 m at 7 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 5·72/3 = 81.66666666666667.


Answer: 500

Prompt: A mass of 5 kg moves in a horizontal circle of radius 1 m at 10 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 5·102/1 = 500.0.


Answer: 18

Prompt: A mass of 4 kg moves in a horizontal circle of radius 2 m at 3 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 4·32/2 = 18.0.


Answer: 113.4

Prompt: A mass of 7 kg moves in a horizontal circle of radius 5 m at 9 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 7·92/5 = 113.4.


Answer: 54

Prompt: A mass of 4 kg moves in a horizontal circle of radius 6 m at 9 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 4·92/6 = 54.0.


Answer: 648

Prompt: A mass of 8 kg moves in a horizontal circle of radius 1 m at 9 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 8·92/1 = 648.0.


Answer: 36

Prompt: A mass of 2 kg moves in a horizontal circle of radius 2 m at 6 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 2·62/2 = 36.0.


Answer: 58.33333333

Prompt: A mass of 7 kg moves in a horizontal circle of radius 3 m at 5 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 7·52/3 = 58.333333333333336.


Answer: 97.2

Prompt: A mass of 6 kg moves in a horizontal circle of radius 5 m at 9 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 6·92/5 = 97.2.


Answer: 64

Prompt: A mass of 6 kg moves in a horizontal circle of radius 6 m at 8 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 6·82/6 = 64.0.


Answer: 49

Prompt: A mass of 6 kg moves in a horizontal circle of radius 6 m at 7 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 6·72/6 = 49.0.


Answer: 96

Prompt: A mass of 8 kg moves in a horizontal circle of radius 3 m at 6 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 8·62/3 = 96.0.


Answer: 125

Prompt: A mass of 5 kg moves in a horizontal circle of radius 1 m at 5 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 5·52/1 = 125.0.


Answer: 196

Prompt: A mass of 4 kg moves in a horizontal circle of radius 1 m at 7 m/s. What is the centripetal force in newtons?

Solution: F = mv2/r = 4·72/1 = 196.0.


Answer: 2.7

Prompt: Two point charges 3e-06 C and 4e-06 C are 0.2 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·1.2e-11/0.04000000000000001 = 2.6999999999999993.


Answer: 0.009

Prompt: Two point charges 1e-06 C and 1e-06 C are 1.0 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·1e-12/1.0 = 0.009.


Answer: 0.288

Prompt: Two point charges 2e-06 C and 4e-06 C are 0.5 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·8e-12/0.25 = 0.288.


Answer: 0.036

Prompt: Two point charges 1e-06 C and 1e-06 C are 0.5 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·1e-12/0.25 = 0.036.


Answer: 0.45

Prompt: Two point charges 1e-06 C and 2e-06 C are 0.2 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·2e-12/0.04000000000000001 = 0.4499999999999999.


Answer: 0.45

Prompt: Two point charges 2e-06 C and 1e-06 C are 0.2 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.


Answer: 1.8

Prompt: Two point charges 2e-06 C and 1e-06 C are 0.1 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·2e-12/0.010000000000000002 = 1.7999999999999996.


Answer: 1.35

Prompt: Two point charges 3e-06 C and 2e-06 C are 0.2 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·6e-12/0.04000000000000001 = 1.3499999999999996.


Answer: 5.4

Prompt: Two point charges 3e-06 C and 2e-06 C are 0.1 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·6e-12/0.010000000000000002 = 5.399999999999999.


Answer: 0.144

Prompt: Two point charges 1e-06 C and 4e-06 C are 0.5 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·4e-12/0.25 = 0.144.


Answer: 0.036

Prompt: Two point charges 2e-06 C and 2e-06 C are 1.0 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·4e-12/1.0 = 0.036.


Answer: 3.6

Prompt: Two point charges 2e-06 C and 2e-06 C are 0.1 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·4e-12/0.010000000000000002 = 3.599999999999999.


Answer: 0.018

Prompt: Two point charges 1e-06 C and 2e-06 C are 1.0 m apart in vacuum. Magnitude of the Coulomb force in newtons? Use k=9.0e9.

Solution: F=k|q1 q2|/r2 = 9e9·2e-12/1.0 = 0.018.


Answer: 3.313035075e-19

Prompt: A photon has frequency 5e+14 Hz. What is its energy in joules? Use h=6.62607015e-34 J·s.

Solution: E=hf=6.62607015e-34·500000000000000.0=3.313035075e-19.


Answer: 3.97564209e-19

Prompt: A photon has frequency 6e+14 Hz. What is its energy in joules? Use h=6.62607015e-34 J·s.


Answer: 4.638249105e-19

Prompt: A photon has frequency 7e+14 Hz. What is its energy in joules? Use h=6.62607015e-34 J·s.

Solution: E=hf=6.62607015e-34·700000000000000.0=4.638249105e-19.


Answer: 6.62607015e-19

Prompt: A photon has frequency 1e+15 Hz. What is its energy in joules? Use h=6.62607015e-34 J·s.

Solution: E=hf=6.62607015e-34·1000000000000000.0=6.62607015e-19.


Answer: 21400.81935

Prompt: Escape speed from a sphere of mass 5.970e+24 kg and radius 1.740e+06 m? Use G=6.6743e-11. Answer in m/s.


Answer: 1692.933236

Prompt: Escape speed from a sphere of mass 7.300e+22 kg and radius 3.400e+06 m? Use G=6.6743e-11. Answer in m/s.


Answer: 11184.9825

Prompt: Escape speed from a sphere of mass 5.970e+24 kg and radius 6.370e+06 m? Use G=6.6743e-11. Answer in m/s.


Answer: 15309.66073

Prompt: Escape speed from a sphere of mass 5.970e+24 kg and radius 3.400e+06 m? Use G=6.6743e-11. Answer in m/s.


Answer: 11213.05026

Prompt: Escape speed from a sphere of mass 6.000e+24 kg and radius 6.370e+06 m? Use G=6.6743e-11. Answer in m/s.


Answer: 2366.489956

Prompt: Escape speed from a sphere of mass 7.300e+22 kg and radius 1.740e+06 m? Use G=6.6743e-11. Answer in m/s.


Answer: 25

Prompt: Beer–Lambert: ε=5000 L·mol−1·cm−1, c=0.005 mol·L−1, l=1.0 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 1

Prompt: Beer–Lambert: ε=2000 L·mol−1·cm−1, c=0.001 mol·L−1, l=0.5 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 10

Prompt: Beer–Lambert: ε=2000 L·mol−1·cm−1, c=0.005 mol·L−1, l=1.0 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 5

Prompt: Beer–Lambert: ε=5000 L·mol−1·cm−1, c=0.002 mol·L−1, l=0.5 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 5

Prompt: Beer–Lambert: ε=5000 L·mol−1·cm−1, c=0.001 mol·L−1, l=1.0 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 50

Prompt: Beer–Lambert: ε=5000 L·mol−1·cm−1, c=0.005 mol·L−1, l=2.0 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 1

Prompt: Beer–Lambert: ε=1000 L·mol−1·cm−1, c=0.002 mol·L−1, l=0.5 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 10

Prompt: Beer–Lambert: ε=5000 L·mol−1·cm−1, c=0.001 mol·L−1, l=2.0 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 0.5

Prompt: Beer–Lambert: ε=1000 L·mol−1·cm−1, c=0.001 mol·L−1, l=0.5 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 20

Prompt: Beer–Lambert: ε=2000 L·mol−1·cm−1, c=0.005 mol·L−1, l=2.0 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 10

Prompt: Beer–Lambert: ε=5000 L·mol−1·cm−1, c=0.002 mol·L−1, l=1.0 cm. What is the absorbance A=εcl (dimensionless)?


Answer: 3.8e-05

Prompt: A capacitor of 2e-06 F is charged to 19 V. What is the stored charge in coulombs?


Answer: 0.00014

Prompt: A capacitor of 1e-05 F is charged to 14 V. What is the stored charge in coulombs?


Answer: 0.00017

Prompt: A capacitor of 1e-05 F is charged to 17 V. What is the stored charge in coulombs?


Answer: 0.000105

Prompt: A capacitor of 5e-06 F is charged to 21 V. What is the stored charge in coulombs?


Answer: 0.0002

Prompt: A capacitor of 1e-05 F is charged to 20 V. What is the stored charge in coulombs?


Answer: 3.5e-05

Prompt: A capacitor of 5e-06 F is charged to 7 V. What is the stored charge in coulombs?


Answer: 6e-05

Prompt: A capacitor of 1e-05 F is charged to 6 V. What is the stored charge in coulombs?


Answer: 7.5e-05

Prompt: A capacitor of 5e-06 F is charged to 15 V. What is the stored charge in coulombs?


Answer: 0.00013

Prompt: A capacitor of 1e-05 F is charged to 13 V. What is the stored charge in coulombs?


Answer: 1.4e-05

Prompt: A capacitor of 2e-06 F is charged to 7 V. What is the stored charge in coulombs?


Answer: 4.6e-05

Prompt: A capacitor of 2e-06 F is charged to 23 V. What is the stored charge in coulombs?


Answer: 5e-05

Prompt: A capacitor of 1e-05 F is charged to 5 V. What is the stored charge in coulombs?


Answer: 167000

Prompt: How much energy in joules is needed for fusion of 0.5 kg of water using L=334000 J/kg (phase change only, no temperature change)?


Answer: 452000

Prompt: How much energy in joules is needed for vaporization of 0.2 kg of water using L=2260000 J/kg (phase change only, no temperature change)?


Answer: 2260000

Prompt: How much energy in joules is needed for vaporization of 1.0 kg of water using L=2260000 J/kg (phase change only, no temperature change)?


Answer: 334000

Prompt: How much energy in joules is needed for fusion of 1.0 kg of water using L=334000 J/kg (phase change only, no temperature change)?


Answer: 1130000

Prompt: How much energy in joules is needed for vaporization of 0.5 kg of water using L=2260000 J/kg (phase change only, no temperature change)?


Answer: 226000

Prompt: How much energy in joules is needed for vaporization of 0.1 kg of water using L=2260000 J/kg (phase change only, no temperature change)?


Answer: 8

Prompt: A planet orbits with period 1.0 year at semi-major axis 1.0 AU. Another body has a=4.0 AU around the same star. Period in years?

Solution: T2∝a3 ⇒ T2=T1·(a2/a1)^(3/2)=8.0.


Answer: 27

Prompt: A planet orbits with period 1.0 year at semi-major axis 1.0 AU. Another body has a=9.0 AU around the same star. Period in years?

Solution: T2∝a3 ⇒ T2=T1·(a2/a1)^(3/2)=27.0.


Answer: 64

Prompt: A planet orbits with period 1.0 year at semi-major axis 1.0 AU. Another body has a=16.0 AU around the same star. Period in years?

Solution: T2∝a3 ⇒ T2=T1·(a2/a1)^(3/2)=64.0.


Answer: 216

Prompt: A constant force of 36 N acts through 12 m at 60° to the displacement. Work in joules?

Solution: W=Fd cosθ=36·12·cos(60°)=216.00000000000006.


Answer: 20

Prompt: A constant force of 8 N acts through 5 m at 60° to the displacement. Work in joules?

Solution: W=Fd cosθ=8·5·cos(60°)=20.000000000000004.


Answer: 80

Prompt: A constant force of 40 N acts through 4 m at 60° to the displacement. Work in joules?

Solution: W=Fd cosθ=40·4·cos(60°)=80.00000000000001.


Answer: 214.5

Prompt: A constant force of 33 N acts through 13 m at 60° to the displacement. Work in joules?

Solution: W=Fd cosθ=33·13·cos(60°)=214.50000000000006.


Answer: 1.469576159e-15

Prompt: A constant force of 12 N acts through 2 m at 90° to the displacement. Work in joules?

Solution: W=Fd cosθ=12·2·cos(90°)=1.4695761589768238e-15.


Answer: 260

Prompt: A constant force of 20 N acts through 13 m at 0° to the displacement. Work in joules?

Solution: W=Fd cosθ=20·13·cos(0°)=260.0.


Answer: 240

Prompt: A constant force of 24 N acts through 10 m at 0° to the displacement. Work in joules?

Solution: W=Fd cosθ=24·10·cos(0°)=240.0.


Answer: 320

Prompt: A constant force of 40 N acts through 8 m at 0° to the displacement. Work in joules?

Solution: W=Fd cosθ=40·8·cos(0°)=320.0.


Answer: 285

Prompt: A constant force of 19 N acts through 15 m at 0° to the displacement. Work in joules?

Solution: W=Fd cosθ=19·15·cos(0°)=285.0.


Answer: 6.245698676e-15

Prompt: A constant force of 34 N acts through 3 m at 90° to the displacement. Work in joules?

Solution: W=Fd cosθ=34·3·cos(90°)=6.245698675651501e-15.


Answer: 7.715274835e-15

Prompt: A constant force of 14 N acts through 9 m at 90° to the displacement. Work in joules?

Solution: W=Fd cosθ=14·9·cos(90°)=7.715274834628325e-15.


Answer: 1.408343819e-14

Prompt: A constant force of 23 N acts through 10 m at 90° to the displacement. Work in joules?

Solution: W=Fd cosθ=23·10·cos(90°)=1.4083438190194563e-14.


Answer: 88

Prompt: A constant force of 22 N acts through 8 m at 60° to the displacement. Work in joules?

Solution: W=Fd cosθ=22·8·cos(60°)=88.00000000000001.


Answer: 87.5

Prompt: A constant force of 35 N acts through 5 m at 60° to the displacement. Work in joules?

Solution: W=Fd cosθ=35·5·cos(60°)=87.50000000000001.


Answer: 1.145044757e-14

Prompt: A constant force of 17 N acts through 11 m at 90° to the displacement. Work in joules?

Solution: W=Fd cosθ=17·11·cos(90°)=1.1450447572027752e-14.


Answer: 1

Molality uses kg of solvent, not solution volume

Prompt: Dissolve 0.5 mol of solute in 0.5 kg of solvent. What is the molality?


Answer: 1

Molality uses kg of solvent, not solution volume

Prompt: Dissolve 1.0 mol of solute in 1.0 kg of solvent. What is the molality?


Answer: 0.5

Molality uses kg of solvent, not solution volume

Prompt: Dissolve 1.0 mol of solute in 2.0 kg of solvent. What is the molality?


Answer: 2

Molality uses kg of solvent, not solution volume

Prompt: Dissolve 1.0 mol of solute in 0.5 kg of solvent. What is the molality?


Answer: 4

Molality uses kg of solvent, not solution volume

Prompt: Dissolve 2.0 mol of solute in 0.5 kg of solvent. What is the molality?


Answer: 1

Molality uses kg of solvent, not solution volume

Prompt: Dissolve 2.0 mol of solute in 2.0 kg of solvent. What is the molality?


Answer: 2

Molality uses kg of solvent, not solution volume

Prompt: Dissolve 2.0 mol of solute in 1.0 kg of solvent. What is the molality?


Answer: 19.47122063

Prompt: Light goes from n=1.0 into n=1.5 at incidence 30°. Refraction angle in degrees?

Solution: n1 sinθ1=n2 sinθ2 ⇒ θ2=19.47122063449069.


Answer: 28.1255057

Prompt: Light goes from n=1.0 into n=1.5 at incidence 45°. Refraction angle in degrees?

Solution: n1 sinθ1=n2 sinθ2 ⇒ θ2=28.12550570205571.


Answer: 0.01591549431

Prompt: An isotropic source radiates 20 W. Intensity at 10 m, in W/m2?

Solution: I=P/(4πr2)=20/(4π·102)=0.015915494309189534.


Answer: 0.9947183943

Prompt: An isotropic source radiates 50 W. Intensity at 2 m, in W/m2?

Solution: I=P/(4πr2)=50/(4π·22)=0.9947183943243458.


Answer: 1.989436789

Prompt: An isotropic source radiates 100 W. Intensity at 2 m, in W/m2?

Solution: I=P/(4πr2)=100/(4π·22)=1.9894367886486917.


Answer: 0.06366197724

Prompt: An isotropic source radiates 20 W. Intensity at 5 m, in W/m2?

Solution: I=P/(4πr2)=20/(4π·52)=0.06366197723675814.


Answer: 0.3978873577

Prompt: An isotropic source radiates 20 W. Intensity at 2 m, in W/m2?

Solution: I=P/(4πr2)=20/(4π·22)=0.3978873577297384.


Answer: 0.3183098862

Prompt: An isotropic source radiates 100 W. Intensity at 5 m, in W/m2?

Solution: I=P/(4πr2)=100/(4π·52)=0.3183098861837907.


Answer: 0.03978873577

Prompt: An isotropic source radiates 50 W. Intensity at 10 m, in W/m2?

Solution: I=P/(4πr2)=50/(4π·102)=0.039788735772973836.


Answer: 0.07957747155

Prompt: An isotropic source radiates 100 W. Intensity at 10 m, in W/m2?

Solution: I=P/(4πr2)=100/(4π·102)=0.07957747154594767.


Answer: 0.03183098862

Prompt: An isotropic source radiates 10 W. Intensity at 5 m, in W/m2?

Solution: I=P/(4πr2)=10/(4π·52)=0.03183098861837907.