HESI A2
HESI A2 Physics Quizlet
1. The specific heat capacity of water is about 2 J/g°C. How much energy would you need to heat 1 kilogram of water by 10°C?
- A. 420 J
- B. 4,200 J
- C. 42,000 J
- D. 420,000 J
Correct answer: C
Rationale: The formula to calculate the energy required to heat a substance is Q = m × c × ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Given that 1 kilogram of water is equal to 1,000 grams, the mass (m) is 1,000 g, the specific heat capacity (c) of water is 4.2 J/g°C (not 2 J/g°C), and the change in temperature (ΔT) is 10°C. Substituting these values into the formula: Q = 1,000 × 4.2 × 10 = 42,000 J. Therefore, the correct energy required to heat 1 kilogram of water by 10°C is 42,000 J. Choices A, B, and D are incorrect as they do not consider the correct specific heat capacity of water or the conversion of mass to grams.
2. How might the energy use of an appliance be expressed?
- A. Power = energy × time
- B. Time + energy = power
- C. Energy = power × time
- D. Energy/power = time
Correct answer: C
Rationale: The energy use of an appliance can be expressed using the formula Energy = Power × Time. In this formula, Energy represents the amount of electricity consumed by the appliance, Power indicates the rate at which the appliance uses electricity (measured in watts), and Time represents the duration for which the appliance is being used (measured in hours). By multiplying the power rating of the appliance by the time it is in use, one can calculate the total energy consumed. Option C is the correct choice because it accurately represents the relationship between power, time, and energy. Choices A, B, and D present incorrect representations of the relationship between energy, power, and time, making them wrong answers.
3. A pitcher throws a 45-g baseball at a velocity of 42 meters per second. What is the ball’s momentum?
- A. 0.189 kg⋅m/s
- B. 1.89 kg⋅m/s
- C. 1.07 kg⋅m/s
- D. 0.93 kg⋅m/s
Correct answer: B
Rationale: Momentum is calculated by multiplying mass (in kg) by velocity (in m/s). The mass of the baseball is 0.045 kg (45 grams converted to kg), and the velocity is 42 m/s. Momentum = 0.045 kg × 42 m/s = 1.89 kg⋅m/s. Therefore, the correct answer is 1.89 kg⋅m/s. Choice A is incorrect as it incorrectly converts the mass from grams to kg. Choice C and D are incorrect due to calculation errors.
4. The speed of sound in dry air at 20°C is 343 m/s. If the wavelength of a sound wave is 5 m, what is its frequency?
- A. 171.5 Hz
- B. 79 Hz
- C. 68.6 Hz
- D. 63.6 Hz
Correct answer: C
Rationale: The speed of sound (v) can be calculated using the formula: v = f × λ, where f is the frequency and λ is the wavelength. Given that the speed of sound is 343 m/s and the wavelength is 5 m, we can rearrange the formula to solve for frequency: f = v / λ = 343 / 5 = 68.6 Hz. Therefore, the correct frequency is 68.6 Hz. Choices A, B, and D are incorrect as they do not result from the correct calculation based on the given values.
5. A 5-kg block is suspended from a spring, causing the spring to stretch 10 cm from equilibrium. What is the spring constant for this spring?
- A. 4.9 N/cm
- B. 9.8 N/cm
- C. 49 N/cm
- D. 50 N/cm
Correct answer: C
Rationale: The spring constant (k) can be calculated using Hooke's Law formula: F = -kx, where F is the force applied, k is the spring constant, and x is the displacement from equilibrium. In this case, the force applied is equal to the weight of the block, F = mg, where m = mass of the block = 5 kg and g = acceleration due to gravity = 9.8 m/s^2. The displacement x = 10 cm = 0.1 m. Substituting the values, we have: 5 kg * 9.8 m/s^2 = k * 0.1 m. Solving for k gives k = 5 * 9.8 / 0.1 = 49 N/m. Therefore, the spring constant for this spring is 49 N/cm. Choice A (4.9 N/cm) is incorrect because it is one decimal place lower than the correct answer. Choice B (9.8 N/cm) is incorrect as it does not account for the correct calculation based on the given information. Choice D (50 N/cm) is incorrect because it is slightly higher than the accurate value obtained through the calculations.
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