Nursing Elites

1. A 110-volt appliance draws 0 amperes. How many watts of power does it require?

• A. 0 watts
• B. 108 watts
• C. 112 watts
• D. 220 watts

Correct answer: A

Rationale: When a 110-volt appliance draws 0 amperes, it means that the power consumption is zero as well. The formula to calculate power is P = V x I, where P is power in watts, V is voltage in volts, and I is current in amperes. Since the current is 0 amperes, the power required by the appliance is also 0 watts. Therefore, the correct answer is 0 watts. Choice B, 108 watts, is incorrect because there is no current drawn. Choice C, 112 watts, and choice D, 220 watts, are incorrect as well since the appliance is not consuming any power when drawing 0 amperes.

2. Given the four wires described here, which would you expect to have the greatest resistance?

• A. 1 km of American wire gauge 1; diameter 7.35 mm
• B. 1 km of American wire gauge 2; diameter 6.54 mm
• C. 1 km of American wire gauge 3; diameter 5.83 mm
• D. 1 km of American wire gauge 4; diameter 5.19 mm

Correct answer: D

Rationale: The wire with the greatest resistance is the one with the smallest diameter, as resistance is inversely proportional to cross-sectional area. Gauge 4 with a 5.19 mm diameter has the smallest diameter and, therefore, the greatest resistance. Choice A, B, and C have larger diameters compared to choice D, so they would have lower resistance values.

3. When calculating an object’s acceleration, what must you do?

• A. Divide the change in time by the velocity.
• B. Multiply the velocity by the time.
• C. Find the difference between the time and velocity.
• D. Divide the change in velocity by the change in time.

Correct answer: D

Rationale: When calculating an object's acceleration, you must divide the change in velocity by the change in time. Acceleration is defined as the rate of change of velocity with respect to time. By determining the ratio of the change in velocity to the change in time, you can ascertain how quickly the velocity of an object is changing, thereby finding its acceleration. Choice A is incorrect because acceleration is not calculated by dividing time by velocity. Choice B is incorrect as it describes multiplying velocity by time, which does not yield acceleration. Choice C is incorrect as finding the difference between time and velocity is not a method to calculate acceleration.

4. How do a scalar quantity and a vector quantity differ?

• A. A scalar quantity has both magnitude and direction, and a vector does not.
• B. A scalar quantity has direction only, and a vector has only magnitude.
• C. A vector has both magnitude and direction, and a scalar quantity has only magnitude.
• D. A vector has only direction, and a scalar quantity has only magnitude.

Correct answer: C

Rationale: The correct answer is C. The main difference between a scalar quantity and a vector quantity lies in the presence of direction. A vector quantity has both magnitude and direction, while a scalar quantity has magnitude only, without any specified direction. Examples of scalar quantities include distance, speed, temperature, and energy, whereas examples of vector quantities include displacement, velocity, force, and acceleration. Choices A, B, and D are incorrect because they incorrectly describe the characteristics of scalar and vector quantities.

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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