Nursing Elites

1. An object with a charge of 4 μC is placed 50 cm from another object with a charge twice as great. What is the magnitude of the resulting repulsive force?

• A. 0.1152 N
• B. 1.152 N
• C. 10^−3 N
• D. 2.5 × 10^−3 N

Correct answer: D

Rationale: The force between two charges is calculated using Coulomb's Law, which states that the force is proportional to the product of the two charges and inversely proportional to the square of the distance between them. Given that one charge is twice as great as the other and the distance between them is 50 cm, we can calculate the repulsive force. The magnitude of the resulting repulsive force is 2.5 × 10^−3 N. Choice A is incorrect as it does not match the calculated value. Choice B is incorrect as it is significantly higher than the correct answer. Choice C is incorrect as it represents 10^−3 N, which is lower than the calculated value.

2. In fluid machinery, pumps are designed to primarily increase the fluid's:

• A. Pressure
• B. Velocity only
• C. Both pressure and velocity
• D. Neither pressure nor velocity

Correct answer: A

Rationale: Pumps in fluid machinery are designed to primarily increase the fluid's pressure. This increase in pressure allows the fluid to flow through the system efficiently and overcome resistance. While pumps can also impact the velocity of the fluid to some extent, their main function is to elevate the pressure to facilitate the movement of the fluid within the system. Choice B is incorrect because pumps do not focus solely on increasing velocity. Choice C is incorrect as while pumps can affect velocity, their primary purpose is to boost pressure. Choice D is incorrect as pumps aim to increase either the pressure, velocity, or both.

3. A plucked guitar string makes 80 vibrations in one second. What is the period?

• A. 0.0125 s
• B. 0.025 s
• C. 0.125 s
• D. 0.25 s

Correct answer: B

Rationale: The period is the time taken for one complete vibration of the guitar string. To find the period, you need to take the reciprocal of the frequency. Since the string makes 80 vibrations in one second, the period is 1/80 = 0.0125 seconds (or 0.025 s). Choice A is incorrect because it is the reciprocal of 80. Choice C is incorrect as it is 10 times the reciprocal of 80. Choice D is incorrect as it is 100 times the reciprocal of 80.

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. A 110-volt hair dryer delivers 1,525 watts of power. How many amperes does it draw?

• A. 167.75 amperes
• B. 1.635 amperes
• C. 1.415 amperes
• D. 13.9 amperes

Correct answer: D

Rationale: To determine the amperes drawn by the hair dryer, we use the formula: Amperes = Watts / Volts. The hair dryer operates at 1,525 watts with 110 volts. Dividing 1,525 watts by 110 volts yields 13.9 amperes. Therefore, the correct answer is 13.9 amperes. Choices A, B, and C are incorrect because they do not result from the correct calculation using the formula.

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