Nursing Elites

1. In terms of electrical conductivity, semiconductors fall between

• A. Conductors and insulators
• B. Conductors and superconductors
• C. Insulators and dielectrics
• D. Superconductors and insulators

Correct answer: A

Rationale: Semiconductors have electrical conductivities that lie between those of conductors (high conductivity) and insulators (low conductivity). This positioning makes choice A, 'Conductors and insulators,' the correct answer. Choice B, 'Conductors and superconductors,' is incorrect because superconductors have perfect conductivity, not intermediate like semiconductors. Choice C, 'Insulators and dielectrics,' is incorrect because dielectrics are a type of insulator, so it doesn't show the progression from high to low conductivity. Choice D, 'Superconductors and insulators,' is incorrect because superconductors have the highest conductivity, opposite to the role of semiconductors.

2. A 3-volt flashlight uses a bulb with 60-ohm resistance. What current flows through the flashlight?

• A. o.05 amp
• B. o.5 amp
• C. 1.8 amp
• D. 18 amp

Correct answer: A

Rationale: : Using Ohm's Law, I = V / R: I = 3 / 60 = 0.05 amp. So, the correct current is 0.05 amp.

3. A 5-kg block is suspended from a spring, causing the spring to stretch 10 cm from equilibrium. What is the spring constant for this spring?

• A. 4.9 N/cm
• B. 9.8 N/cm
• C. 49 N/cm
• D. 50 N/cm

Correct answer: C

Rationale: The spring constant (k) can be calculated using Hooke's Law formula: F = -kx, where F is the force applied, k is the spring constant, and x is the displacement from equilibrium. In this case, the force applied is equal to the weight of the block, F = mg, where m = mass of the block = 5 kg and g = acceleration due to gravity = 9.8 m/s^2. The displacement x = 10 cm = 0.1 m. Substituting the values, we have: 5 kg * 9.8 m/s^2 = k * 0.1 m. Solving for k gives k = 5 * 9.8 / 0.1 = 49 N/m. Therefore, the spring constant for this spring is 49 N/cm. Choice A (4.9 N/cm) is incorrect because it is one decimal place lower than the correct answer. Choice B (9.8 N/cm) is incorrect as it does not account for the correct calculation based on the given information. Choice D (50 N/cm) is incorrect because it is slightly higher than the accurate value obtained through the calculations.

4. A 2,000-kg car travels at 15 m/s. For a 1,500-kg car traveling at 15 m/s to generate the same momentum, what would need to happen?

• A. It would need to accelerate to 20 m/s.
• B. It would need to add 500 kg in mass.
• C. Both A and B
• D. Either A or B

Correct answer: A

Rationale: Momentum is calculated as the product of mass and velocity. Since momentum is conserved in the absence of external forces, for the 1,500-kg car to generate the same momentum as the 2,000-kg car at 15 m/s, it would need to increase its velocity to compensate for the difference in mass. Accelerating to 20 m/s would achieve this without needing to change the mass of the car. Choice B is incorrect because adding mass is not necessary to match momentum in this scenario.

5. A 5-cm candle is placed 20 cm away from a concave mirror with a focal length of 15 cm. About what is the image height of the candle in the mirror?

• A. 30.5 cm
• B. 15.625 cm
• C. −15 cm
• D. −30.5 cm

Correct answer: B

Rationale: The magnification formula for a mirror is given by M = -f / (f - d), where f is the focal length of the mirror, and d is the object distance from the mirror. Using the mirror equation and magnification formula, the image height is found to be negative because it is inverted. Plugging in the values (f = 15 cm, d = 20 cm) into the formula gives M = -15 / (15 - 20) = -15 / -5 = 3. The negative sign indicates that the image is inverted. The image height is then calculated by multiplying the magnification by the object height: 3 * 5 cm = 15 cm. Therefore, the correct image height is approximately -15 cm. Choice A (30.5 cm) and Choice D (-30.5 cm) are incorrect as they do not consider the inversion of the image. Choice C (-15 cm) is also incorrect because it neglects the negative sign, which indicates the inversion of the image.

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