an object with a charge of 3 c is placed 30 cm from another object with a charge of 2 c what is the magnitude of the resulting force between the objec
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HESI A2

HESI A2 Physics Practice Test

1. An object with a charge of 3 μC is placed 30 cm from another object with a charge of 2 μC. What is the magnitude of the resulting force between the objects?

Correct answer: B

Rationale: To find the magnitude of the resulting force between two charges, we use Coulomb's Law: F = k × (|q1 × q2|) / r² Where: F is the force k is Coulomb’s constant (8.99 × 10⁹ N·m²/C²) q1 and q2 are the charges r is the distance between the charges Plugging in the values: F = (8.99 × 10⁹) × (3 × 10⁻⁶) × (2 × 10⁻⁶) / (0.3)² = 0.18 N. Therefore, the magnitude of the resulting force is 0.18 N.

2. A hummingbird’s wings beat at 25 beats per second. What is the period of the wing beating in seconds?

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.

3. For a compressible fluid subjected to rapid pressure changes, sound wave propagation becomes important. The speed of sound (c) depends on the fluid's:

Correct answer: C

Rationale: In a compressible fluid, the speed of sound (c) depends on both the fluid's density (ρ) and Bulk modulus. Density affects the compressibility of the fluid, while Bulk modulus represents the fluid's resistance to compression and plays a crucial role in determining the speed of sound in a compressible medium. Viscosity and surface tension do not directly impact the speed of sound in a compressible fluid subjected to rapid pressure changes. Therefore, the correct answer is C.

4. An object with a charge of 4 μC is placed 50 cm from another object with a charge twice as great. What is the magnitude of the resulting repulsive force?

Correct answer: D

Rationale: The force between two charges is calculated using Coulomb's Law, which states that the force is proportional to the product of the two charges and inversely proportional to the square of the distance between them. Given that one charge is twice as great as the other and the distance between them is 50 cm, we can calculate the repulsive force. The magnitude of the resulting repulsive force is 2.5 × 10^−3 N. Choice A is incorrect as it does not match the calculated value. Choice B is incorrect as it is significantly higher than the correct answer. Choice C is incorrect as it represents 10^−3 N, which is lower than the calculated value.

5. An object has a constant velocity of 50 m/s and travels for 10 s. What is the acceleration of the object?

Correct answer: A

Rationale: The acceleration of an object is defined as the rate of change of its velocity. When an object has a constant velocity, it means there is no change in its speed or direction. In this case, the object maintains a constant velocity of 50 m/s for 10 seconds, which implies that there is no change in velocity. Therefore, the acceleration of the object is 0 m/s² as there is no acceleration or deceleration happening. Choices B, C, and D are incorrect because acceleration is the change in velocity over time, and in this scenario of constant velocity, the acceleration is 0 m/s².

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