in open channel flow a critical property is the free surface which refers to the

HESI A2

HESI A2 Physics Practice Test

1. 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.

2. Why are boats more buoyant in salt water than in fresh water?

A. Salt decreases the mass of the boats.
B. Salt increases the volume of the water.
C. Salt affects the density of the boats.
D. Salt increases the density of the water.

Correct answer: D

Rationale: Salt increases the density of water, making saltwater more buoyant than freshwater. The higher density of saltwater provides more lift to a boat, enabling it to float more easily compared to in freshwater. Choice A is incorrect because salt does not affect the mass of the boats. Choice B is incorrect as salt does not increase the volume of water. Choice C is incorrect since salt affects the density of water, not the boats themselves. Therefore, the correct answer is that salt increases the density of the water, resulting in boats being more buoyant in salt water than in fresh water.

3. What is the electric field inside a hollow conductor with a net charge?

A. Remains constant
B. Decreases
C. Zero
D. Becomes unpredictable

Correct answer: C

Rationale: The correct answer is C: Zero. According to Gauss’s Law, the electric field inside a hollow conductor (a conductor with no charge inside but a net charge on its surface) is zero. The charges reside on the outer surface of the conductor, causing the electric field inside to cancel out. Choices A, B, and D are incorrect because the electric field inside a hollow conductor with a net charge is not constant, does not decrease, and does not become unpredictable; it is zero due to the distribution of charges on its surface.

4. Which conclusion can be drawn from Ohm’s law?

A. Voltage and current are inversely proportional when resistance is constant.
B. The ratio of the potential difference between the ends of a conductor to current is a constant, R.
C. Voltage is the amount of charge that passes through a point per second.
D. Power (P) can be calculated by multiplying current (I) by voltage (V).

Correct answer: B

Rationale: Ohm's law states that the ratio of the potential difference (voltage) between the ends of a conductor to the current flowing through it is a constant. Mathematically, this is represented as V = I x R, where V is voltage, I is current, and R is the constant resistance. Therefore, the correct conclusion that can be drawn from Ohm's law is that the ratio of the potential difference between the ends of a conductor to current is a constant, denoted as R. This relationship is fundamental to understanding the behavior of electrical circuits and the effect of resistance on voltage and current. Choice A is incorrect because Ohm's law actually states that voltage and current are directly proportional when resistance is constant. Choice C is incorrect because voltage is not the amount of charge that passes through a point per second; rather, it is the electric potential energy per unit charge. Choice D is incorrect because although power (P) can be calculated by multiplying current (I) by voltage (V), this is not a conclusion directly drawn from Ohm's law.

5. What is the kinetic energy of a 500-kg wagon moving at 10 m/s?

A. 50 J
B. 250 J
C. 2.5 × 10^4 J
D. 5.0 × 10^5 J

Correct answer: C

Rationale: The formula for calculating kinetic energy is KE = 0.5 × mass × velocity². Given the mass of the wagon is 500 kg and the velocity is 10 m/s, we can substitute these values into the formula: KE = 0.5 × 500 kg × (10 m/s)² = 0.5 × 500 kg × 100 m²/s² = 25,000 J or 2.5 × 10⁴ J. Therefore, the kinetic energy of the 500-kg wagon moving at 10 m/s is 2.5 × 10⁴ J. Choice A (50 J) is incorrect because it is too low; Choice B (250 J) is incorrect as it does not match the correct calculation; Choice D (5.0 × 10^5 J) is incorrect as it is too high. The correct answer is C (2.5 × 10^4 J).