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. For steady, incompressible flow through a pipe, the mass flow rate (ṁ) is related to the fluid density (ρ), cross-sectional area (A), and average velocity (v) via the continuity equation:

• A. ṁ cannot be determined without additional information
• B. ṁ = ρvA
• C. Bernoulli's principle is solely applicable here
• D. The equation of state for the specific fluid is required

Correct answer: B

Rationale: The continuity equation for steady, incompressible flow states that the mass flow rate is the product of the fluid's density, velocity, and cross-sectional area. Hence, ṁ = ρvA. Choice A is incorrect because the mass flow rate can be determined using the given formula. Choice C is incorrect as Bernoulli's principle does not directly relate to the mass flow rate calculation. Choice D is incorrect as the equation of state is not needed to calculate the mass flow rate in this scenario.

3. What force was applied to the object that was moved if 100 N⋅m of work is done over 20 m?

• A. 5 N
• B. 80 N
• C. 120 N
• D. 2,000 N

Correct answer: A

Rationale: Work is calculated using the formula Work = Force x Distance. Given that 100 N⋅m of work is done over 20 m, we can rearrange the formula to solve for Force. Force = Work / Distance. Plugging in the values, we get Force = 100 N⋅m / 20 m = 5 N. Therefore, the force applied to the object that was moved is 5 N. Choice B (80 N) is incorrect because it doesn't match the calculated force of 5 N. Choice C (120 N) is incorrect as it is higher than the calculated force. Choice D (2,000 N) is incorrect as it is significantly higher than the correct force of 5 N.

4. Two objects attract each other with a gravitational force of 12 units. If you double the distance between the objects, what is the new force of attraction between the two?

• A. 3 units
• B. 6 units
• C. 24 units
• D. 48 units

Correct answer: A

Rationale: The gravitational force between two objects is inversely proportional to the square of the distance between them. If the distance is doubled, the force will be reduced to 1/4 of the original force. Therefore, the new force of attraction between the two objects will be 12 units / 4 = 3 units. Choice A is correct because doubling the distance reduces the force to 1/4 of the original value. Choices B, C, and D are incorrect as they do not consider the inverse square relationship between distance and gravitational force.

5. 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.

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