HESI A2
HESI A2 Physics Practice Test
1. What does Coulomb’s law relate to?
- A. electrostatic interaction
- B. rigid body motion
- C. heat conduction
- D. universal gravitation
Correct answer: A
Rationale: Coulomb's law is a fundamental principle in physics that deals with the electrostatic interaction between charged particles. It states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This law is crucial in understanding and predicting the behavior of electrically charged objects. Choices B, C, and D are incorrect because Coulomb's law specifically focuses on electrostatic interactions between charges, not rigid body motion, heat conduction, or universal gravitation.
2. The speed of sound in dry air at 20°C is 343 m/s. If the wavelength of a sound wave is 5 m, what is its frequency?
- A. 171.5 Hz
- B. 79 Hz
- C. 68.6 Hz
- D. 63.6 Hz
Correct answer: C
Rationale: The speed of sound (v) can be calculated using the formula: v = f × λ, where f is the frequency and λ is the wavelength. Given that the speed of sound is 343 m/s and the wavelength is 5 m, we can rearrange the formula to solve for frequency: f = v / λ = 343 / 5 = 68.6 Hz. Therefore, the correct frequency is 68.6 Hz. Choices A, B, and D are incorrect as they do not result from the correct calculation based on the given values.
3. 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.
4. 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.
5. Amanda uses 100 N of force to push a lawnmower around her lawn. If she mows 20 rows measuring 30 meters each, how much work does she do?
- A. 3,000 N⋅m
- B. 6,000 N⋅m
- C. 60,000 N⋅m
- D. The answer cannot be determined from the information given.
Correct answer: C
Rationale: The work done by Amanda pushing the lawnmower is calculated by multiplying the force applied (100 N) by the distance over which the force is applied (the total distance mowed). Since Amanda mows 20 rows, each measuring 30 meters, the total distance mowed is 20 rows x 30 meters/row = 600 meters. Therefore, the work done is 100 N x 600 m = 60,000 N⋅m. Option A and B are incorrect as they do not account for the total distance mowed. Option D is incorrect as the work done can be accurately calculated based on the information provided.
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