Nursing Elites

1. 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.

2. In a U-tube manometer, a fluid is used to measure pressure differences. When one side is connected to a pressurized system, the fluid level on that side will:

• A. Remain the same
• B. Decrease
• C. Increase
• D. Depend on the type of fluid used

Correct answer: B

Rationale: In a U-tube manometer, the side connected to a pressurized system will experience a decrease in fluid level due to the pressure exerted by the system. This pressure forces the fluid down, causing the fluid level to decrease. Therefore, choice B is correct. Choices A and C are incorrect because the fluid level will not remain the same or increase when connected to a pressurized system. Choice D is incorrect as the type of fluid used does not determine the direction of the fluid movement in response to pressure.

3. A circular running track has a circumference of 2,500 meters. What is the radius of the track?

• A. 1,000 m
• B. 400 m
• C. 25 m
• D. 12 m

Correct answer: B

Rationale: The radius of a circular track can be calculated using the formula: Circumference = 2 × π × radius. Given that the circumference of the track is 2,500 m, we can plug this into the formula and solve for the radius: 2,500 = 2 × π × radius. Dividing both sides by 2π gives: radius = 2,500 / (2 × 3.1416) ≈ 397.89 m. Therefore, the closest answer is 400 m, making option B the correct choice. Option A (1,000 m) is too large, option C (25 m) is too small, and option D (12 m) is significantly smaller than the calculated radius.

4. What is the SI unit for quantifying the transfer of energy due to an applied force?

• A. Newton (N)
• B. Meter per second (m/s)
• C. Joule (J)
• D. Kilogram (kg)

Correct answer: C

Rationale: The correct answer is C: Joule (J). The joule is the SI unit used to quantify the transfer of energy due to an applied force. It is defined as the work done when a force of one newton is applied over a distance of one meter. Newton (N) is the unit of force, not energy transfer. Meter per second (m/s) is the unit of speed, not energy transfer. Kilogram (kg) is the unit of mass, not energy transfer. Therefore, the correct unit for quantifying the transfer of energy due to an applied force is the joule (J).

5. Certain non-Newtonian fluids exhibit shear thickening behavior. In this case, the fluid's viscosity:

• A. Remains constant with increasing shear rate
• B. Decreases with increasing shear rate (shear thinning)
• C. Increases with increasing shear rate
• D. Depends solely on the applied pressure

Correct answer: C

Rationale: When a non-Newtonian fluid exhibits shear thickening behavior, its viscosity increases with increasing shear rate. This means that as more force is applied to the fluid, its resistance to flow also increases, resulting in a higher viscosity. This phenomenon is opposite to shear thinning, where viscosity decreases with increasing shear rate. Therefore, in the case of shear thickening behavior, the correct answer is that the fluid's viscosity increases with increasing shear rate. Choices A, B, and D are incorrect because shear thickening behavior specifically involves an increase in viscosity with increasing shear rate, not remaining constant, decreasing, or depending on applied pressure.

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