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. Two 5-ohm resistors are placed in series and wired into a 100-V power supply. What current flows through this circuit?

• A. 2 A
• B. 10 A
• C. 20 A
• D. 50 A

Correct answer: B

Rationale: In a series circuit, the total resistance is the sum of the individual resistances. Therefore, the total resistance in this circuit is 5 ohms + 5 ohms = 10 ohms. Using Ohm's Law (V = I × R), we can find the current (I) by dividing the voltage (V) by the total resistance (R). I = V / R = 100 V / 10 ohms = 10 A. Choice A (2 A) is incorrect because it does not account for the total resistance of the circuit. Choice C (20 A) and Choice D (50 A) are also incorrect as they provide values that are not consistent with the calculations based on the given values in the question.

3. If a force of 12 kg stretches a spring by 3 cm, how far will the spring stretch when a force of 30 kg is applied?

• A. 6 cm
• B. 7.5 cm
• C. 9 cm
• D. 10.5 cm

Correct answer: B

Rationale: The extension of a spring is directly proportional to the force applied. In this case, the force increases from 12 kg to 30 kg, which is a 2.5 times increase. Therefore, the extension of the spring will also increase by 2.5 times. Given that the spring stretches 3 cm with a force of 12 kg, multiplying 3 cm by 2.5 gives us the extension of the spring when a force of 30 kg is applied, which equals 7.5 cm. Therefore, the correct answer is 7.5 cm. Choice A, 6 cm, is incorrect because it does not account for the proportional increase in force. Choice C, 9 cm, and Choice D, 10.5 cm, are incorrect as they overestimate the extension of the spring by not considering the direct proportionality between force and extension.

4. A concave mirror with a focal length of 2 cm forms a real image of an object at an image distance of 6 cm. What is the object's distance from the mirror?

• A. 3 cm
• B. 6 cm
• C. 12 cm
• D. 30 cm

Correct answer: B

Rationale: The mirror formula, 1/f = 1/do + 1/di, can be used to solve for the object distance. Given that the focal length (f) is 2 cm and the image distance (di) is 6 cm, we can substitute these values into the formula to find the object distance. Plugging in f = 2 cm and di = 6 cm into the formula gives us 1/2 = 1/do + 1/6. Solving for do, we get do = 6 cm. Therefore, the object's distance from the mirror is 6 cm. Choice A (3 cm), Choice C (12 cm), and Choice D (30 cm) are incorrect distances as the correct object distance is determined to be 6 cm.

5. A 10-kg object moving at 5 m/s has an impulse acted on it causing the velocity to change to 15 m/s. What was the impulse that was applied to the object?

• A. 10 kg⋅m/s
• B. 15 kg⋅m/s
• C. 20 kg⋅m/s
• D. 100 kg⋅m/s

Correct answer: D

Rationale: Impulse is the change in momentum of an object. The initial momentum is calculated as 10 kg × 5 m/s = 50 kg⋅m/s, and the final momentum is 10 kg × 15 m/s = 150 kg⋅m/s. The change in momentum (impulse) is 150 kg⋅m/s - 50 kg⋅m/s = 100 kg⋅m/s. Therefore, the impulse applied to the object is 100 kg⋅m/s. Choices A, B, and C are incorrect because they do not reflect the correct calculation of the impulse based on the change in momentum of the object.

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