when two identical charged spheres both positively charged are brought close together the electrostatic force between them will be
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HESI A2

HESI A2 Physics

1. When two identical charged spheres, both positively charged, are brought close together, the electrostatic force between them will be:

Correct answer: D

Rationale: When two positively charged spheres are brought close together, they will experience a repulsive force due to their like charges. The electrostatic force causes the spheres to repel each other, making the correct answer D: Strongly repulsive. The force is not dependent on the material of the spheres, and the force is definitely not zero, as like charges repel. Choice A is incorrect as like charges do not attract each other. Choice C is incorrect as like charges repel, not attract.

2. An object with a charge of 3 μC is placed 30 cm from another object with a charge of 2 μC. What is the magnitude of the resulting force between the objects?

Correct answer: B

Rationale: To find the magnitude of the resulting force between two charges, we use Coulomb's Law: F = k × (|q1 × q2|) / r² Where: F is the force k is Coulomb’s constant (8.99 × 10⁹ N·m²/C²) q1 and q2 are the charges r is the distance between the charges Plugging in the values: F = (8.99 × 10⁹) × (3 × 10⁻⁶) × (2 × 10⁻⁶) / (0.3)² = 0.18 N. Therefore, the magnitude of the resulting force is 0.18 N.

3. The drag force (F_d) experienced by an object moving through a fluid depends on:

Correct answer: B

Rationale: The drag force experienced by an object moving through a fluid depends on multiple factors, including the object's shape, size, velocity, and the fluid's properties such as viscosity and density. Choices A, C, and D are incorrect because drag force is not solely determined by the object's shape and size, depth of submersion, or buoyant force acting on the object. The primary factors affecting drag force are the fluid properties and the object's velocity. Therefore, the correct answer is B.

4. A caterpillar starts moving at a rate of 14 in/hr. After 15 minutes, it is moving at a rate of 20 in/hr. What is the caterpillar’s rate of acceleration?

Correct answer: C

Rationale: Acceleration is the change in velocity over time. The change in velocity for the caterpillar is 20 in/hr - 14 in/hr = 6 in/hr. Since this change occurred over 15 minutes (or 0.25 hours), the acceleration can be calculated as (6 in/hr) / (0.25 hr) = 24 in/hr². Therefore, the caterpillar's rate of acceleration is 24 in/hr², which corresponds to choice C. Choice A, 6 in/hr², is incorrect as it does not account for the time factor and the correct calculation. Choice B, 12 in/hr², is incorrect as it doubles the correct acceleration value. Choice D, 280 in/hr², is significantly higher than the correct value, indicating a calculation error.

5. What force was applied to the object that was moved if 100 N⋅m of work is done over 20 m?

Correct answer: A

Rationale: Work is calculated using the formula Work = Force x Distance. Given that 100 N⋅m of work is done over 20 m, we can rearrange the formula to solve for Force. Force = Work / Distance. Plugging in the values, we get Force = 100 N⋅m / 20 m = 5 N. Therefore, the force applied to the object that was moved is 5 N. Choice B (80 N) is incorrect because it doesn't match the calculated force of 5 N. Choice C (120 N) is incorrect as it is higher than the calculated force. Choice D (2,000 N) is incorrect as it is significantly higher than the correct force of 5 N.

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