Nursing Elites

1. 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.

2. What is the diameter of a loop if its radius is 6 meters?

• A. 6 m
• B. 12 m
• C. 18 m
• D. 36 m

Correct answer: B

Rationale: The diameter of a loop is calculated by multiplying the radius by 2. Since the radius is 6 meters, the diameter is 6 × 2 = 12 meters. Therefore, the correct answer is 12 meters. Choice A (6 m) is the radius, not the diameter. Choices C (18 m) and D (36 m) are incorrect as they do not reflect the correct calculation for determining the diameter of a loop.

3. A solenoid is a long, tightly wound coil of wire that acts like a bar magnet when current flows through it. The magnetic field lines inside a solenoid are most similar to the field lines around:

• A. A single straight current-carrying wire
• B. A horseshoe magnet
• C. A permanent bar magnet
• D. A flat sheet conductor

Correct answer: C

Rationale: The magnetic field lines inside a solenoid resemble the field lines around a permanent bar magnet. Both a solenoid and a bar magnet have north and south poles, resulting in a similar pattern of magnetic field lines. A single straight current-carrying wire produces a different field pattern because it has no coil structure like a solenoid. A horseshoe magnet has a unique field shape due to its pole arrangement, different from the uniform field pattern of a solenoid. A flat sheet conductor does not exhibit the same magnetic field characteristics as a solenoid, as it lacks the coil shape and alignment of a solenoid's magnetic field.

4. What is the kinetic energy of a 500-kg wagon moving at 10 m/s?

• A. 50 J
• B. 250 J
• C. 2.5 × 10^4 J
• D. 5.0 × 10^5 J

Correct answer: C

Rationale: The formula for calculating kinetic energy is KE = 0.5 × mass × velocity². Given the mass of the wagon is 500 kg and the velocity is 10 m/s, we can substitute these values into the formula: KE = 0.5 × 500 kg × (10 m/s)² = 0.5 × 500 kg × 100 m²/s² = 25,000 J or 2.5 × 10⁴ J. Therefore, the kinetic energy of the 500-kg wagon moving at 10 m/s is 2.5 × 10⁴ J. Choice A (50 J) is incorrect because it is too low; Choice B (250 J) is incorrect as it does not match the correct calculation; Choice D (5.0 × 10^5 J) is incorrect as it is too high. The correct answer is C (2.5 × 10^4 J).

5. As a car is traveling on the highway, its speed drops from 60 mph to 30 mph. What happens to its kinetic energy?

• A. Its energy is halved.
• B. Its energy is doubled.
• C. Its energy is quadrupled.
• D. Its energy is divided by four.

Correct answer: A

Rationale: The correct answer is A. Kinetic energy is proportional to the square of the velocity. When the speed drops from 60 mph to 30 mph, the kinetic energy is halved. Choice B is incorrect because halving the speed results in halving the kinetic energy, not doubling it. Choice C is incorrect because quadrupling the kinetic energy would require increasing the speed fourfold, not halving it. Choice D is incorrect because dividing the energy by four would imply a different relationship between speed and kinetic energy, which is not the case.

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