Nursing Elites

1. A 25-cm spring stretches to 28 cm when a force of 12 N is applied. What would its length be if that force were doubled?

• A. 31 cm
• B. 40 cm
• C. 50 cm
• D. 56 cm

Correct answer: A

Rationale: When the 12 N force stretches the spring from 25 cm to 28 cm, it causes a length increase of 28 cm - 25 cm = 3 cm. Therefore, each newton of applied force causes an extension of 3 cm / 12 N = 0.25 cm/N. If the force is doubled to 24 N, the spring would extend by 24 N × 0.25 cm/N = 6 cm more than its original length of 25 cm. Thus, the new length of the spring would be 25 cm + 6 cm = 31 cm. Choice A, 31 cm, is the correct answer as calculated. Choices B, C, and D are incorrect as they do not consider the relationship between force and extension in the spring, leading to incorrect calculations of the new length.

2. 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.

3. The frequency of an alternating current (AC) refers to the number of times it changes direction per unit time. This is measured in:

• A. Hertz
• B. Amperes
• C. Volts
• D. Ohms

Correct answer: A

Rationale: The frequency of an alternating current (AC) is measured in Hertz (Hz), which denotes the number of times the current changes direction per unit time. Hertz is the unit for frequency, while amperes measure current, volts measure voltage, and ohms measure resistance. Therefore, the correct answer is Hertz (Hz). Choices B, C, and D are incorrect because amperes measure current intensity, volts measure voltage potential, and ohms measure resistance, not the frequency of an alternating current.

4. Two balloons with charges of 5 μC each are placed 25 cm apart. What is the magnitude of the resulting repulsive force between them?

• A. 0.18 N
• B. 1.8 N
• C. 10−3 N
• D. 5 × 10−3 N

Correct answer: B

Rationale: To find the repulsive force between the two charges, we use Coulomb's law: F = k(q1 * q2) / r^2. Here, k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2), q1 and q2 are the charges (5 μC each), and r is the distance between the charges (25 cm = 0.25 m). Substituting these values into the formula: F = (8.99 x 10^9 Nm^2/C^2)(5 x 10^-6 C)(5 x 10^-6 C) / (0.25 m)^2. Calculating this gives F = 1.8 N. Therefore, the magnitude of the resulting repulsive force between the two balloons is 1.8 N. Choice A, C, and D are incorrect as they do not correctly calculate the force using Coulomb's law.

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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