Nursing Elites

1. A wave moves through its medium at 20 m/s with a wavelength of 4 m. What is the frequency of the wave?

• A. 5 s−1
• B. 16 s−1
• C. 24 s−1
• D. 80 s−1

Correct answer: C

Rationale: The formula to calculate the frequency of a wave is given by:

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

• A. 0 m/s²
• B. 5 m/s²
• C. 60 m/s²
• D. 500 m/s²

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.

3. What is the net force acting on the car?

• A. 450 N
• B. 700 N
• C. 1,500 N
• D. 6,300 N

Correct answer: C

Rationale: To determine the net force acting on an object, we need to consider the sum of the forces acting in the same direction and subtract the forces acting in the opposite direction. In this scenario, there is a force of 4,200 N to the right and a force of 2,700 N to the left. By subtracting the leftward force from the rightward force (4,200 N - 2,700 N), we find that the net force acting on the car is 1,500 N to the right. Therefore, choice C, 1,500 N, is the correct answer. Choice A, 450 N, is too small as it does not account for the total forces involved. Choice B, 700 N, is also incorrect as it is not the result of the correct mathematical operation on the given forces. Choice D, 6,300 N, is too large and does not align with the calculation based on the forces provided.

4. Why does potential energy increase as particles approach each other?

• A. Attractive forces increase.
• B. Attractive forces decrease.
• C. Repulsive forces increase.
• D. Repulsive forces decrease.

Correct answer: C

Rationale: The correct answer is C: Repulsive forces increase. As particles approach each other, the distance between them decreases, causing the repulsive forces between the particles to increase. This increase in repulsive forces leads to an increase in potential energy as the particles resist being pushed closer together. Choices A and B are incorrect because attractive forces do not increase or decrease in this scenario. Choice D is incorrect because repulsive forces actually increase as particles get closer, leading to a rise in potential energy.

5. 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?

• A. 0.1152 N
• B. 1.152 N
• C. 10^−3 N
• D. 2.5 × 10^−3 N

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.

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