Nursing Elites

1. What force was applied to the object that was moved if 100 Nâ‹…m of work is done over 20 m?

• A. 5 N
• B. 80 N
• C. 120 N
• D. 2,000 N

Correct answer: A

Rationale: Work is calculated using the formula Work = Force x Distance. Given that 100 Nâ‹…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 Nâ‹…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.

2. For a compressible fluid subjected to rapid pressure changes, sound wave propagation becomes important. The speed of sound (c) depends on the fluid's:

• A. Density (ρ) only
• B. Viscosity (μ) only
• C. Density (ρ) and Bulk modulus
• D. Density (ρ) and Surface tension (γ)

Correct answer: C

Rationale: In a compressible fluid, the speed of sound (c) depends on both the fluid's density (ρ) and Bulk modulus. Density affects the compressibility of the fluid, while Bulk modulus represents the fluid's resistance to compression and plays a crucial role in determining the speed of sound in a compressible medium. Viscosity and surface tension do not directly impact the speed of sound in a compressible fluid subjected to rapid pressure changes. Therefore, the correct answer is C.

3. 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.

4. An object with a charge of 4 μC is placed 1 meter from another object with a charge of 2 μC. What is the magnitude of the resulting force between the objects?

• A. 0.04 N
• B. 0.072 N
• C. 80 N
• D. 8 × 10−6 N

Correct answer: A

Rationale: To find the magnitude of the resulting force between two charges, we can use Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula for Coulomb's law is: F = k × (|q1 × q2| / r²), where F is the force, k is the Coulomb constant, q1 and q2 are the charges, and r is the distance between the charges. Substituting the given values into the formula: F = (9 × 10⁹ N·m²/C²) × ((4 × 10⁻⁶ C) × (2 × 10⁻⁶ C) / (1 m)²) = 0.04 N. Therefore, the magnitude of the resulting force between the objects is 0.04 N.

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

• A. Slightly attractive
• B. Zero
• C. Strongly attractive
• D. Strongly repulsive

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.

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