Nursing Elites

1. 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.

2. Which substance would be most affected by a change in temperature?

• A. Liquid nitrogen
• B. Salt crystals
• C. Hydrogen gas
• D. Iron filings

Correct answer: C

Rationale: Hydrogen gas would be most affected by a change in temperature because gases have a greater expansion or contraction in volume with changes in temperature compared to liquids or solids. When the temperature of hydrogen gas increases, its molecules gain kinetic energy and move faster, causing the gas to expand and its volume to increase. Conversely, when the temperature decreases, the gas molecules lose kinetic energy and move slower, leading to a decrease in volume. This property makes hydrogen gas highly sensitive to temperature changes compared to liquid nitrogen, salt crystals, or iron filings. Liquid nitrogen, salt crystals, and iron filings are less affected by temperature changes because their particles are closer together and have lower kinetic energy, resulting in minimal volume changes with temperature fluctuations.

3. A common example of a shear-thinning (non-Newtonian) fluid is:

• A. Water
• B. Ketchup
• C. Air
• D. Alcohol

Correct answer: B

Rationale: The correct answer is B: Ketchup. Shear-thinning fluids become less viscous under stress. Ketchup is an example of a shear-thinning fluid because its viscosity decreases when it is shaken or squeezed, allowing it to flow more easily. Choice A, Water, is a Newtonian fluid with a constant viscosity regardless of stress. Choice C, Air, is also a Newtonian fluid. Choice D, Alcohol, does not exhibit shear-thinning behavior; it typically has a constant viscosity as well.

4. Ocean waves build during a storm until there is a vertical distance from the high point to the low point of 6 meters and a horizontal distance of 9 meters between adjacent crests. The waves hit the shore every 5 seconds. What is the speed of the waves?

• A. 1.2 m/s
• B. 1.8 m/s
• C. 2.0 m/s
• D. 2.4 m/s

Correct answer: B

Rationale: To find the speed of the waves, we use the formula: speed = wavelength / period. The wavelength is the horizontal distance between adjacent crests, which is 9 meters in this case. The period is the time it takes for one wave to pass a fixed point, given as 5 seconds. Therefore, speed = 9 meters / 5 seconds = 1.8 m/s. Choice A (1.2 m/s) is incorrect because it miscalculates the speed. Choice C (2.0 m/s) and Choice D (2.4 m/s) are incorrect as they do not correctly calculate the speed using the provided data.

5. 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.

Similar Questions