Nursing Elites

1. What is the primary factor responsible for generating lift on an airplane wing?

• A. Propulsion force generated by the engines
• B. Buoyant forces acting on the entire aircraft
• C. Drag reduction achieved through streamlining
• D. Application of Bernoulli's principle to the airfoil's shape

Correct answer: D

Rationale: The primary factor responsible for generating lift on an airplane wing is the application of Bernoulli's principle. This principle states that the air moving over the curved top surface of the wing has to travel faster, leading to reduced pressure above the wing and creating lift. Engines provide thrust for propulsion, not lift. Buoyant forces are more relevant to lighter-than-air aircraft like balloons or airships, not airplanes. While drag reduction through streamlining is important for efficiency, it is not the primary factor in lift generation. Therefore, the correct answer is D.

2. Two balloons with charges of 5 μC each are placed 25 cm apart. What is the magnitude of the resulting repulsive force between them?

• A. 0.18 N
• B. 1.8 N
• C. 10−3 N
• D. 5 × 10−3 N

Correct answer: B

Rationale: To find the repulsive force between the two charges, we use Coulomb's law: F = k(q1 * q2) / r^2. Here, k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2), q1 and q2 are the charges (5 μC each), and r is the distance between the charges (25 cm = 0.25 m). Substituting these values into the formula: F = (8.99 x 10^9 Nm^2/C^2)(5 x 10^-6 C)(5 x 10^-6 C) / (0.25 m)^2. Calculating this gives F = 1.8 N. Therefore, the magnitude of the resulting repulsive force between the two balloons is 1.8 N. Choice A, C, and D are incorrect as they do not correctly calculate the force using Coulomb's law.

3. 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.

4. When light travels from air into a denser medium like glass, its speed:

• A. Increases
• B. Decreases
• C. Remains constant
• D. Becomes unpredictable

Correct answer: B

Rationale: When light travels from air into a denser medium like glass, its speed decreases. This is because the higher refractive index of the denser medium causes light to slow down as it propagates through the medium. Choice A is incorrect because the speed of light decreases in a denser medium. Choice C is incorrect because the speed of light changes when it enters a denser medium. Choice D is incorrect because the change in speed is predictable based on the refractive index of the medium.

5. 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.

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