Nursing Elites

1. Why are boats more buoyant in salt water than in fresh water?

• A. Salt decreases the mass of the boats.
• B. Salt increases the volume of the water.
• C. Salt affects the density of the boats.
• D. Salt increases the density of the water.

Correct answer: D

Rationale: Salt increases the density of water, making saltwater more buoyant than freshwater. The higher density of saltwater provides more lift to a boat, enabling it to float more easily compared to in freshwater. Choice A is incorrect because salt does not affect the mass of the boats. Choice B is incorrect as salt does not increase the volume of water. Choice C is incorrect since salt affects the density of water, not the boats themselves. Therefore, the correct answer is that salt increases the density of the water, resulting in boats being more buoyant in salt water than in fresh water.

2. Which of these substances is most compressible?

• A. Gold
• B. Water
• C. Mercury
• D. Methane

Correct answer: D

Rationale: Methane, a gas at room temperature and pressure, is the most compressible substance among the options provided. Gases are generally more compressible compared to liquids and solids because their particles have more space between them, allowing for greater compression when pressure is applied. Gold, water, and mercury, being solid and liquid substances, respectively, have particles arranged closely together, making them less compressible. Therefore, the correct answer is Methane.

3. A 10-kg object moving at 5 m/s has an impulse acted on it causing the velocity to change to 15 m/s. What was the impulse that was applied to the object?

• A. 10 kgâ‹…m/s
• B. 15 kgâ‹…m/s
• C. 20 kgâ‹…m/s
• D. 100 kgâ‹…m/s

Correct answer: D

Rationale: Impulse is the change in momentum of an object. The initial momentum is calculated as 10 kg × 5 m/s = 50 kg⋅m/s, and the final momentum is 10 kg × 15 m/s = 150 kg⋅m/s. The change in momentum (impulse) is 150 kg⋅m/s - 50 kg⋅m/s = 100 kg⋅m/s. Therefore, the impulse applied to the object is 100 kg⋅m/s. Choices A, B, and C are incorrect because they do not reflect the correct calculation of the impulse based on the change in momentum of the object.

4. A spring has a spring constant of 20 N/m. How much force is needed to compress the spring from 40 cm to 30 cm?

• A. 200 N
• B. 80 N
• C. 5 N
• D. 2 N

Correct answer: D

Rationale: The change in length of the spring is 40 cm - 30 cm = 10 cm = 0.10 m. The force required to compress or stretch a spring is given by Hooke's Law: F = k × x, where F is the force, k is the spring constant (20 N/m in this case), and x is the change in length (0.10 m). Substituting the values into the formula: F = 20 N/m × 0.10 m = 2 N. Therefore, the correct answer is 2 N. Choice A (200 N) is incorrect because it miscalculates the force. Choice B (80 N) is incorrect as it does not apply Hooke's Law correctly. Choice C (5 N) is incorrect as it underestimates the force required.

5. A constant force is exerted on a stationary object. In this scenario, work is:

• A. Performed
• B. Not performed
• C. Partially performed
• D. Inconclusive without further information

Correct answer: B

Rationale: Work is only done when a force causes displacement. Since the object is stationary, no displacement occurs, and therefore, no work is performed. Choice A is incorrect because work requires both force and displacement. Choice C is incorrect as there is no partial work - work is either done or not done. Choice D is incorrect as the scenario provided is clear - the object is stationary, so no work is being performed.

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