Nursing Elites

1. A caterpillar starts moving at a rate of 14 in/hr. After 15 minutes, it is moving at a rate of 20 in/hr. What is the caterpillar’s rate of acceleration?

• A. 6 in/hr²
• B. 12 in/hr²
• C. 24 in/hr²
• D. 280 in/hr²

Correct answer: C

Rationale: Acceleration is the change in velocity over time. The change in velocity for the caterpillar is 20 in/hr - 14 in/hr = 6 in/hr. Since this change occurred over 15 minutes (or 0.25 hours), the acceleration can be calculated as (6 in/hr) / (0.25 hr) = 24 in/hr². Therefore, the caterpillar's rate of acceleration is 24 in/hr², which corresponds to choice C. Choice A, 6 in/hr², is incorrect as it does not account for the time factor and the correct calculation. Choice B, 12 in/hr², is incorrect as it doubles the correct acceleration value. Choice D, 280 in/hr², is significantly higher than the correct value, indicating a calculation error.

2. When a crane hoists a massive object at a constant velocity compared to lifting the same object gradually, the work done by the crane is:

• A. Less
• B. More
• C. Identical
• D. Dependent on the object's mass

Correct answer: C

Rationale: The work done by the crane is identical in both scenarios. Work is defined as the force applied over a distance. Since the force needed to lift the object is equal to its weight and the displacement is the same, the work done is identical, whether the object is lifted gradually or at a constant velocity. Choice A is incorrect because the work done is the same in both cases. Choice B is incorrect as well since the work done does not increase. Choice D is incorrect as the mass of the object does not affect the work done by the crane in this scenario.

3. A 60-watt lightbulb is powered by a 110-volt power source. What is the current being drawn?

• A. 0.55 amperes
• B. 1.83 amperes
• C. 50 amperes
• D. 6,600 amperes

Correct answer: A

Rationale: To calculate the current being drawn, use the formula I = P / V, where I is the current, P is the power in watts, and V is the voltage. Substituting the given values, I = 60 / 110 ≈ 0.55 amperes. Therefore, the current being drawn by the 60-watt lightbulb is approximately 0.55 amperes. Choice B, 1.83 amperes, is incorrect as it does not match the calculated value. Choices C and D, 50 amperes and 6,600 amperes, are significantly higher values and do not align with the expected current draw of a 60-watt lightbulb powered by a 110-volt source.

4. In a parallel circuit, the ___________ through each component is the same.

• A. current
• B. voltage
• C. resistance
• D. wattage

Correct answer: A

Rationale: In a parallel circuit, the current through each component is the same. This is because the components in a parallel circuit are connected across the same voltage source, so they all experience the same voltage across their terminals. The total current entering the parallel circuit is then split up among the various components, but the current through each component remains the same as the total current. Choices B, C, and D are incorrect. In a parallel circuit, voltage across each component may vary, resistance may differ, and wattage is related to power, not the equality of current through each component.

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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