Nursing Elites

1. If a force of 12 kg stretches a spring by 3 cm, how far will the spring stretch when a force of 30 kg is applied?

• A. 6 cm
• B. 7.5 cm
• C. 9 cm
• D. 10.5 cm

Correct answer: B

Rationale: The extension of a spring is directly proportional to the force applied. In this case, the force increases from 12 kg to 30 kg, which is a 2.5 times increase. Therefore, the extension of the spring will also increase by 2.5 times. Given that the spring stretches 3 cm with a force of 12 kg, multiplying 3 cm by 2.5 gives us the extension of the spring when a force of 30 kg is applied, which equals 7.5 cm. Therefore, the correct answer is 7.5 cm. Choice A, 6 cm, is incorrect because it does not account for the proportional increase in force. Choice C, 9 cm, and Choice D, 10.5 cm, are incorrect as they overestimate the extension of the spring by not considering the direct proportionality between force and extension.

2. A hummingbird’s wings beat at 25 beats per second. What is the period of the wing beating in seconds?

• A. 0.04 s
• B. 0.25 s
• C. 0.4 s
• D. 4 s

Correct answer: A

Rationale: The period represents the time for one complete cycle of the wing beating. To calculate the period, you take the reciprocal of the frequency. In this case, with the wings beating at 25 beats per second, the period is 1/25, which equals 0.04 seconds. Therefore, choice A, 0.04 seconds, is correct. Choices B, C, and D are incorrect because they do not reflect the correct calculation of the period based on the given frequency of 25 beats per second.

3. A 5-cm candle is placed 20 cm away from a concave mirror with a focal length of 10 cm. What is the image distance of the candle?

• A. 20 cm
• B. 40 cm
• C. 60 cm
• D. 75 cm

Correct answer: C

Rationale: To find the image distance of the candle, we use the mirror formula: 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. In this case, the focal length f = 10 cm and the object distance do = 20 cm. Substituting these values into the formula gives us 1/10 = 1/20 + 1/di. Solving for di, we get di = 60 cm. Therefore, the image distance of the candle is 60 cm. Choice A (20 cm) is incorrect because it represents the object distance, not the image distance. Choice B (40 cm) is incorrect as it does not consider the mirror formula calculation. Choice D (75 cm) is incorrect as it does not match the correct calculation based on the mirror formula.

4. 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.

5. If the force acting on an object is doubled, how does its acceleration change?

• A. It remains the same.
• B. It is halved.
• C. It is doubled.
• D. It is eliminated.

Correct answer: C

Rationale: According to Newton's second law of motion, the acceleration of an object is directly proportional to the force acting on it. Therefore, if the force acting on an object is doubled, its acceleration will also double. This relationship is expressed by the equation F = ma, where F is the force, m is the mass of the object, and a is the acceleration. When the force (F) is doubled, the acceleration (a) will also double, assuming the mass remains constant. Choice A is incorrect because acceleration changes with a change in force. Choice B is incorrect because acceleration and force are directly proportional. Choice D is incorrect because increasing the force acting on an object does not eliminate its acceleration; instead, it results in an increase in acceleration, as per Newton's second law.

Similar Questions