Nursing Elites

1. Given the four wires described here, which would you expect to have the greatest resistance?

• A. 1 km of American wire gauge 1; diameter 7.35 mm
• B. 1 km of American wire gauge 2; diameter 6.54 mm
• C. 1 km of American wire gauge 3; diameter 5.83 mm
• D. 1 km of American wire gauge 4; diameter 5.19 mm

Correct answer: D

Rationale: The wire with the greatest resistance is the one with the smallest diameter, as resistance is inversely proportional to cross-sectional area. Gauge 4 with a 5.19 mm diameter has the smallest diameter and, therefore, the greatest resistance. Choice A, B, and C have larger diameters compared to choice D, so they would have lower resistance values.

2. What is the SI unit for quantifying the transfer of energy due to an applied force?

• A. Newton (N)
• B. Meter per second (m/s)
• C. Joule (J)
• D. Kilogram (kg)

Correct answer: C

Rationale: The correct answer is C: Joule (J). The joule is the SI unit used to quantify the transfer of energy due to an applied force. It is defined as the work done when a force of one newton is applied over a distance of one meter. Newton (N) is the unit of force, not energy transfer. Meter per second (m/s) is the unit of speed, not energy transfer. Kilogram (kg) is the unit of mass, not energy transfer. Therefore, the correct unit for quantifying the transfer of energy due to an applied force is the joule (J).

3. The amount of energy lost in a circuit due to electrical resistance is dissipated in the form of:

• A. Light
• B. Sound
• C. Heat
• D. Mechanical work

Correct answer: C

Rationale: When electrical current flows through a circuit with resistance, energy is lost in the form of heat due to the resistance encountered by the electrons. This dissipation of energy as heat is a common phenomenon in electrical circuits and is known as Joule heating. Therefore, the correct answer is 'Heat.' Light, sound, and mechanical work are not typical forms in which energy is lost due to electrical resistance. Light is not a direct result of energy dissipation in electrical circuits, sound is not a form of energy dissipation in this context, and mechanical work pertains to the application of physical force and not the dissipation of energy due to resistance.

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 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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