Nursing Elites

1. What is the main difference between a reversible and irreversible process in thermodynamics?

• A. Reversible processes involve heat transfer, while irreversible processes do not.
• B. Reversible processes occur instantaneously, while irreversible processes take time.
• C. Reversible processes can be run in both directions with the same outcome, while irreversible processes cannot.
• D. Reversible processes violate the first law of thermodynamics.

Correct answer: C

Rationale: A reversible process is an idealized process that can be reversed without leaving any change in either the system or the surroundings. In contrast, irreversible processes cannot be reversed and often involve entropy production or dissipation. Choice A is incorrect because both reversible and irreversible processes can involve heat transfer. Choice B is incorrect as the speed of a process does not determine its reversibility. Choice D is incorrect because reversible processes do not violate the first law of thermodynamics; they comply with it by maintaining a balance between energy inputs and outputs. Therefore, the correct answer is C, as it accurately captures the main difference between reversible and irreversible processes in thermodynamics.

2. A closed system undergoes a cyclic process, returning to its initial state. What can be said about the net work done (Wnet) by the system over the entire cycle?

• A. Wnet is always positive.
• B. Wnet is always negative.
• C. Wnet can be positive, negative, or zero.
• D. Wnet is equal to the total heat transferred into the system (dQ ≠ 0 for a cycle).

Correct answer: C

Rationale: For a closed system undergoing a cyclic process and returning to its initial state, the net work done (Wnet) over the entire cycle can be positive, negative, or zero. This is because the work done is determined by the area enclosed by the cycle on a P-V diagram, and this area can be above, below, or intersecting the zero work axis, leading to positive, negative, or zero net work done. Choice A is incorrect because Wnet is not always positive; it depends on the specific path taken on the P-V diagram. Choice B is incorrect as Wnet is not always negative; it varies based on the enclosed area. Choice D is incorrect because Wnet is not necessarily equal to the total heat transferred into the system; it depends on the specifics of the cycle and is not a direct relationship.

3. Two balloons with charges of 5 μC each are placed 25 cm apart. What is the magnitude of the resulting repulsive force between them?

• A. 0.18 N
• B. 1.8 N
• C. 10−3 N
• D. 5 × 10−3 N

Correct answer: B

Rationale: To find the repulsive force between the two charges, we use Coulomb's law: F = k(q1 * q2) / r^2. Here, k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2), q1 and q2 are the charges (5 μC each), and r is the distance between the charges (25 cm = 0.25 m). Substituting these values into the formula: F = (8.99 x 10^9 Nm^2/C^2)(5 x 10^-6 C)(5 x 10^-6 C) / (0.25 m)^2. Calculating this gives F = 1.8 N. Therefore, the magnitude of the resulting repulsive force between the two balloons is 1.8 N. Choice A, C, and D are incorrect as they do not correctly calculate the force using Coulomb's law.

4. If a 5-kg ball is moving at 5 m/s, what is its momentum?

• A. 10 kgâ‹…m/s
• B. 16.2 km/h
• C. 24.75 kgâ‹…m/s
• D. 25 kgâ‹…m/s

Correct answer: D

Rationale: The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the mass of the ball is 5 kg and its velocity is 5 m/s. Therefore, the momentum of the ball is 5 kg × 5 m/s = 25 kg⋅m/s. Choice A (10 kg⋅m/s) is incorrect as it does not account for both mass and velocity. Choice B (16.2 km/h) is incorrect as it provides a speed in a different unit without considering mass. Choice C (24.75 kg⋅m/s) is incorrect as it does not correctly calculate the momentum based on the given mass and velocity.

5. What is the kinetic energy of a 500-kg wagon moving at 10 m/s?

• A. 50 J
• B. 250 J
• C. 2.5 × 10^4 J
• D. 5.0 × 10^5 J

Correct answer: C

Rationale: The formula for calculating kinetic energy is KE = 0.5 × mass × velocity². Given the mass of the wagon is 500 kg and the velocity is 10 m/s, we can substitute these values into the formula: KE = 0.5 × 500 kg × (10 m/s)² = 0.5 × 500 kg × 100 m²/s² = 25,000 J or 2.5 × 10⁴ J. Therefore, the kinetic energy of the 500-kg wagon moving at 10 m/s is 2.5 × 10⁴ J. Choice A (50 J) is incorrect because it is too low; Choice B (250 J) is incorrect as it does not match the correct calculation; Choice D (5.0 × 10^5 J) is incorrect as it is too high. The correct answer is C (2.5 × 10^4 J).

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