HESI A2
HESI A2 Physics
1. During an isothermal (constant temperature) expansion, what is the work done by the gas on the surroundings?
- A. Positive and equal to the change in internal energy.
- B. Zero.
- C. Negative and equal to the change in internal energy.
- D. Positive and greater than the change in internal energy.
Correct answer: D
Rationale: In an isothermal expansion, the temperature remains constant, meaning there is no change in internal energy. However, the gas still does work on the surroundings as it expands, and this work is positive. Since internal energy does not change, the correct answer is D, 'Positive and greater than the change in internal energy.' Choice A is incorrect because the work done is not equal to the change in internal energy. Choice B is incorrect as work is done during the expansion. Choice C is incorrect since the work done is not negative during an isothermal expansion.
2. According to the Clausius inequality, for a cyclic process involving heat transfer between a system and its surroundings at a single constant temperature (T), the following inequality must hold true:
- A. There is no relationship between heat transfer and temperature in a cyclic process.
- B. ∫ dQ/T ≥ 0
- C. ∫ Q/T = constant
- D. ∫ dQ/T ≤ 0
Correct answer: D
Rationale: The Clausius inequality states that for a cyclic process involving heat transfer at a single constant temperature, the integral of heat transfer divided by temperature (∫ dQ/T) must be less than or equal to zero. This inequality reflects the irreversibility of natural processes. Choice A is incorrect as there is a direct relationship between heat transfer and temperature in the Clausius inequality. Choice B is incorrect because the integral of dQ/T must be less than or equal to zero, not greater than or equal to zero. Choice C is incorrect because the integral of Q/T is not a constant in a cyclic process involving heat transfer at a single constant temperature.
3. Faraday's law of electromagnetic induction states that a changing magnetic field in a conductor induces a/an:
- A. Increase in resistance
- B. Electromotive force
- C. Static electric charge
- D. Decrease in capacitance
Correct answer: B
Rationale: Faraday's law of electromagnetic induction states that a changing magnetic field induces an electromotive force in a conductor. This electromotive force is responsible for generating electricity in power plants and various electrical devices. The induced current is a result of the changing magnetic field, not an increase in resistance (choice A), static electric charge (choice C), or a decrease in capacitance (choice D). Hence, the correct answer is B.
4. A 2,000-kg car travels at 15 m/s. For a 1,500-kg car traveling at 15 m/s to generate the same momentum, what would need to happen?
- A. It would need to accelerate to 20 m/s.
- B. It would need to add 500 kg in mass.
- C. Both A and B
- D. Either A or B
Correct answer: A
Rationale: Momentum is calculated as the product of mass and velocity. Since momentum is conserved in the absence of external forces, for the 1,500-kg car to generate the same momentum as the 2,000-kg car at 15 m/s, it would need to increase its velocity to compensate for the difference in mass. Accelerating to 20 m/s would achieve this without needing to change the mass of the car. Choice B is incorrect because adding mass is not necessary to match momentum in this scenario.
5. Jack stands in front of a plane mirror. If he is 5 feet away from the mirror, how far away from Jack is his image?
- A. 2.5 feet
- B. 3 feet
- C. 4.5 feet
- D. 5 feet
Correct answer: D
Rationale: When Jack stands in front of a plane mirror, his image appears the same distance behind the mirror as Jack is in front of it. Therefore, if Jack is 5 feet away from the mirror, his image will also appear 5 feet behind the mirror. The total distance from Jack to his image is the sum of these distances, which equals 10 feet. Choices A, B, and C are incorrect because the image distance is not half of the total distance but the same as the object's distance from the mirror.
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