Nursing Elites

1. 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.

2. An object with a charge of 3 μC is placed 30 cm from another object with a charge of 2 μC. What is the magnitude of the resulting force between the objects?

• A. 0.6 N
• B. 0.18 N
• C. 180 N
• D. 9 × 10−12 N

Correct answer: B

Rationale: To find the magnitude of the resulting force between two charges, we use Coulomb's Law: F = k × (|q1 × q2|) / r² Where: F is the force k is Coulomb’s constant (8.99 × 10⁹ N·m²/C²) q1 and q2 are the charges r is the distance between the charges Plugging in the values: F = (8.99 × 10⁹) × (3 × 10⁻⁶) × (2 × 10⁻⁶) / (0.3)² = 0.18 N. Therefore, the magnitude of the resulting force is 0.18 N.

3. How do you determine the velocity of a wave?

• A. Multiply the frequency by the wavelength.
• B. Add the frequency and the wavelength.
• C. Subtract the wavelength from the frequency.
• D. Divide the wavelength by the frequency.

Correct answer: A

Rationale: The velocity of a wave can be determined by multiplying the frequency of the wave by the wavelength. This relationship is given by the formula: velocity = frequency × wavelength. By multiplying the frequency by the wavelength, you can calculate the speed at which the wave is traveling. This formula is derived from the basic wave equation v = f × λ, where v represents velocity, f is frequency, and λ is wavelength. Therefore, to find the velocity of a wave, one must multiply its frequency by its wavelength. Choices B, C, and D are incorrect. Adding, subtracting, or dividing the frequency and wavelength does not yield the correct calculation for wave velocity. The correct formula for determining wave velocity is to multiply the frequency by the wavelength.

4. A 120-volt heat lamp draws 25 amps of current. What is the lamp’s resistance?

• A. 96 ohms
• B. 104 ohms
• C. 150 ohms
• D. 4.8 ohms

Correct answer: D

Rationale: To find the resistance of the lamp, we use Ohm’s Law, which states that resistance (R) is equal to voltage (V) divided by current (I), expressed as: R = V / I. Given that the voltage (V) is 120 volts and the current (I) is 25 amps, we substitute these values into the formula: R = 120 V / 25 A = 4.8 ohms. Therefore, the resistance of the lamp is 4.8 ohms. Choice A, 96 ohms, is incorrect as it is not the result of the correct calculation. Choice B, 104 ohms, is incorrect as it does not match the calculated resistance. Choice C, 150 ohms, is incorrect as it is not the correct resistance value obtained through the calculation.

5. Household alternating current typically has a frequency of 60 Hz. Which statement is true?

• A. The circuit is suitable for lighting 60-watt bulbs.
• B. Circuits in the home may carry a current of 60 amperes.
• C. The expected voltage drop is 60 volts per meter.
• D. Electrons complete a cycle 60 times per second.

Correct answer: D

Rationale: The correct answer is D. Electrons complete a cycle 60 times per second when the frequency of the current is 60 Hz. This frequency indicates that the current changes direction 60 times per second, causing the electrons to complete a full cycle back and forth through the circuit at the same rate. Choice A is incorrect because the power rating of a bulb (in watts) is not directly related to the frequency of the current. Choice B is incorrect as typical household circuits do not carry currents as high as 60 amperes. Choice C is incorrect as the expected voltage drop is not measured in volts per meter for household alternating current circuits.

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