Nursing Elites

1. 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.

2. In a U-tube manometer, a fluid is used to measure pressure differences. When one side is connected to a pressurized system, the fluid level on that side will:

• A. Remain the same
• B. Decrease
• C. Increase
• D. Depend on the type of fluid used

Correct answer: B

Rationale: In a U-tube manometer, the side connected to a pressurized system will experience a decrease in fluid level due to the pressure exerted by the system. This pressure forces the fluid down, causing the fluid level to decrease. Therefore, choice B is correct. Choices A and C are incorrect because the fluid level will not remain the same or increase when connected to a pressurized system. Choice D is incorrect as the type of fluid used does not determine the direction of the fluid movement in response to pressure.

3. 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.

4. Two 5-ohm resistors are placed in series and wired into a 100-V power supply. What current flows through this circuit?

• A. 2 A
• B. 10 A
• C. 20 A
• D. 50 A

Correct answer: B

Rationale: In a series circuit, the total resistance is the sum of the individual resistances. Therefore, the total resistance in this circuit is 5 ohms + 5 ohms = 10 ohms. Using Ohm's Law (V = I × R), we can find the current (I) by dividing the voltage (V) by the total resistance (R). I = V / R = 100 V / 10 ohms = 10 A. Choice A (2 A) is incorrect because it does not account for the total resistance of the circuit. Choice C (20 A) and Choice D (50 A) are also incorrect as they provide values that are not consistent with the calculations based on the given values in the question.

5. 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.

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