Nursing Elites

1. Why are boats more buoyant in salt water than in fresh water?

• A. Salt decreases the mass of the boats.
• B. Salt increases the volume of the water.
• C. Salt affects the density of the boats.
• D. Salt increases the density of the water.

Correct answer: D

Rationale: Salt increases the density of water, making saltwater more buoyant than freshwater. The higher density of saltwater provides more lift to a boat, enabling it to float more easily compared to in freshwater. Choice A is incorrect because salt does not affect the mass of the boats. Choice B is incorrect as salt does not increase the volume of water. Choice C is incorrect since salt affects the density of water, not the boats themselves. Therefore, the correct answer is that salt increases the density of the water, resulting in boats being more buoyant in salt water than in fresh water.

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?

• A. 0.6 N
• B. 0.18 N
• C. 180 N
• D. 9 × 10−12 N

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. A 120-volt heat lamp draws 25 amps of current. What is the lamp’s resistance?

• A. 96 ohms
• B. 104 ohms
• C. 150 ohms
• D. 4.8 ohms

Correct answer: D

Rationale: To find the resistance of the lamp, we use Ohm’s Law, which states that resistance (R) is equal to voltage (V) divided by current (I), expressed as: R = V / I. Given that the voltage (V) is 120 volts and the current (I) is 25 amps, we substitute these values into the formula: R = 120 V / 25 A = 4.8 ohms. Therefore, the resistance of the lamp is 4.8 ohms. Choice A, 96 ohms, is incorrect as it is not the result of the correct calculation. Choice B, 104 ohms, is incorrect as it does not match the calculated resistance. Choice C, 150 ohms, is incorrect as it is not the correct resistance value obtained through the calculation.

4. A 60-watt lightbulb is powered by a 110-volt power source. What is the current being drawn?

• A. 0.55 amperes
• B. 1.83 amperes
• C. 50 amperes
• D. 6,600 amperes

Correct answer: A

Rationale: To calculate the current being drawn, use the formula I = P / V, where I is the current, P is the power in watts, and V is the voltage. Substituting the given values, I = 60 / 110 ≈ 0.55 amperes. Therefore, the current being drawn by the 60-watt lightbulb is approximately 0.55 amperes. Choice B, 1.83 amperes, is incorrect as it does not match the calculated value. Choices C and D, 50 amperes and 6,600 amperes, are significantly higher values and do not align with the expected current draw of a 60-watt lightbulb powered by a 110-volt source.

5. 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.

Similar Questions