a plucked guitar string makes 80 vibrations in one second what is the period

HESI A2

HESI Exams Quizlet Physics

1. 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.

2. A system undergoes an isobaric process (constant pressure). In this process, the work done (W) by the system is:

A. Zero, if the volume change (ΔV) is zero.
B. Positive and equal to the pressure multiplied by the volume change (W = PΔV).
C. Negative and equal to the pressure multiplied by the volume change.
D. Independent of the pressure or volume change.

Correct answer: B

Rationale: In an isobaric process (constant pressure), the work done is given by the formula W = PΔV, where P is the pressure and ΔV is the change in volume. If the volume does not change, the work done is zero, not negative. Choice A is incorrect as it states the work done is zero when the volume change is zero, which is the correct condition for zero work. Choice C is incorrect as it incorrectly suggests that the work done is negative in an isobaric process. Choice D is incorrect as the work done in an isobaric process is indeed dependent on the volume change and pressure.

3. If a 5-kg ball is moving at 5 m/s, what is its momentum?

A. 10 kg⋅m/s
B. 16.2 km/h
C. 24.75 kg⋅m/s
D. 25 kg⋅m/s

Correct answer: D

Rationale: The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the mass of the ball is 5 kg and its velocity is 5 m/s. Therefore, the momentum of the ball is 5 kg × 5 m/s = 25 kg⋅m/s. Choice A (10 kg⋅m/s) is incorrect as it does not account for both mass and velocity. Choice B (16.2 km/h) is incorrect as it provides a speed in a different unit without considering mass. Choice C (24.75 kg⋅m/s) is incorrect as it does not correctly calculate the momentum based on the given mass and velocity.

4. A common example of a shear-thinning (non-Newtonian) fluid is:

A. Water
B. Ketchup
C. Air
D. Alcohol

Correct answer: B

Rationale: The correct answer is B: Ketchup. Shear-thinning fluids become less viscous under stress. Ketchup is an example of a shear-thinning fluid because its viscosity decreases when it is shaken or squeezed, allowing it to flow more easily. Choice A, Water, is a Newtonian fluid with a constant viscosity regardless of stress. Choice C, Air, is also a Newtonian fluid. Choice D, Alcohol, does not exhibit shear-thinning behavior; it typically has a constant viscosity as well.

5. In open-channel flow, a critical property is the free surface, which refers to the:

A. Interface between the liquid and the container walls
B. Interface between the liquid and a surrounding gas
C. Bottom of the channel
D. Region of highest velocity within the liquid

Correct answer: B

Rationale: The free surface in open-channel flow refers to the interface between the liquid and the surrounding gas, typically the atmosphere. This interface is critical as it determines the boundary between the liquid flow and the open environment. Option A is incorrect as it refers to the liquid-container wall interface, not the free surface. Option C is incorrect because it represents the bottom of the channel, not the free surface. Option D is incorrect as it describes the region of highest velocity within the liquid, not the free surface. Therefore, the correct choice is B.