Nursing Elites

1. 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.

2. 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.

3. A 5-kg block is suspended from a spring, causing the spring to stretch 10 cm from equilibrium. What is the spring constant for this spring?

• A. 4.9 N/cm
• B. 9.8 N/cm
• C. 49 N/cm
• D. 50 N/cm

Correct answer: C

Rationale: The spring constant (k) can be calculated using Hooke's Law formula: F = -kx, where F is the force applied, k is the spring constant, and x is the displacement from equilibrium. In this case, the force applied is equal to the weight of the block, F = mg, where m = mass of the block = 5 kg and g = acceleration due to gravity = 9.8 m/s^2. The displacement x = 10 cm = 0.1 m. Substituting the values, we have: 5 kg * 9.8 m/s^2 = k * 0.1 m. Solving for k gives k = 5 * 9.8 / 0.1 = 49 N/m. Therefore, the spring constant for this spring is 49 N/cm. Choice A (4.9 N/cm) is incorrect because it is one decimal place lower than the correct answer. Choice B (9.8 N/cm) is incorrect as it does not account for the correct calculation based on the given information. Choice D (50 N/cm) is incorrect because it is slightly higher than the accurate value obtained through the calculations.

4. Viscosity, μ, is a transport property of a fluid that reflects its:

• A. Inertia
• B. Resistance to flow
• C. Compressibility
• D. Buoyancy generation

Correct answer: B

Rationale: Viscosity refers to a fluid's resistance to flow. A fluid with high viscosity (like honey) flows slowly, while a fluid with low viscosity (like water) flows more easily. It is a measure of internal friction in the fluid. Choice A, 'Inertia,' is incorrect as inertia is the tendency of an object to resist changes in its state of motion. Choice C, 'Compressibility,' is incorrect as it refers to the ability of a fluid to be compressed. Choice D, 'Buoyancy generation,' is incorrect as it relates to the upward force exerted by a fluid that opposes the weight of an immersed object.

5. Which object below has the same density?

• A. A block with a mass of 6.5 grams and a volume of 16.25 cm3
• B. A block with a mass of 80 grams and a volume of 32 cm3
• C. A block with a mass of 48 grams and a volume of 22 cm3
• D. A block with a mass of 100 grams and a volume of 250 cm3

Correct answer: A

Rationale: Density is calculated by dividing the mass of an object by its volume. The density of object A is 6.5 g / 16.25 cm3 = 0.4 g/cm3. The density of object B is 80 g / 32 cm3 = 2.5 g/cm3. The density of object C is 48 g / 22 cm3 = 2.18 g/cm3. The density of object D is 100 g / 250 cm3 = 0.4 g/cm3. Objects A and D have the same density of 0.4 g/cm3. Therefore, the correct answer is A as it has the same density as object D, making them the only objects with a density of 0.4 g/cm3.

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