Nursing Elites

1. The amount of energy lost in a circuit due to electrical resistance is dissipated in the form of:

• A. Light
• B. Sound
• C. Heat
• D. Mechanical work

Correct answer: C

Rationale: When electrical current flows through a circuit with resistance, energy is lost in the form of heat due to the resistance encountered by the electrons. This dissipation of energy as heat is a common phenomenon in electrical circuits and is known as Joule heating. Therefore, the correct answer is 'Heat.' Light, sound, and mechanical work are not typical forms in which energy is lost due to electrical resistance. Light is not a direct result of energy dissipation in electrical circuits, sound is not a form of energy dissipation in this context, and mechanical work pertains to the application of physical force and not the dissipation of energy due to resistance.

2. 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.

3. 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.

4. During an isothermal (constant temperature) expansion, what is the work done by the gas on the surroundings?

• A. Positive and equal to the change in internal energy.
• B. Zero.
• C. Negative and equal to the change in internal energy.
• D. Positive and greater than the change in internal energy.

Correct answer: D

Rationale: In an isothermal expansion, the temperature remains constant, meaning there is no change in internal energy. However, the gas still does work on the surroundings as it expands, and this work is positive. Since internal energy does not change, the correct answer is D, 'Positive and greater than the change in internal energy.' Choice A is incorrect because the work done is not equal to the change in internal energy. Choice B is incorrect as work is done during the expansion. Choice C is incorrect since the work done is not negative during an isothermal expansion.

5. 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.

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