Nursing Elites

1. 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.

2. If a wave has a frequency of 60 hertz, which of the following is true?

• A. It completes one cycle per minute.
• B. It measures 60 m from crest to crest.
• C. It completes 60 cycles per second.
• D. It measures 60 m from crest to trough.

Correct answer: C

Rationale: The frequency of a wave is the number of cycles it completes in one second. A wave with a frequency of 60 hertz completes 60 cycles per second. Therefore, choice C is correct. Choice A is incorrect because a frequency of 60 hertz means 60 cycles per second, not per minute. Choice B is incorrect as the frequency of the wave does not determine the distance from crest to crest. Choice D is also incorrect as the frequency does not relate to the distance from crest to trough.

3. 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.

4. A concave mirror with a focal length of 2 cm forms a real image of an object at an image distance of 6 cm. What is the object's distance from the mirror?

• A. 3 cm
• B. 6 cm
• C. 12 cm
• D. 30 cm

Correct answer: B

Rationale: The mirror formula, 1/f = 1/do + 1/di, can be used to solve for the object distance. Given that the focal length (f) is 2 cm and the image distance (di) is 6 cm, we can substitute these values into the formula to find the object distance. Plugging in f = 2 cm and di = 6 cm into the formula gives us 1/2 = 1/do + 1/6. Solving for do, we get do = 6 cm. Therefore, the object's distance from the mirror is 6 cm. Choice A (3 cm), Choice C (12 cm), and Choice D (30 cm) are incorrect distances as the correct object distance is determined to be 6 cm.

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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