Nursing Elites

1. A hummingbird’s wings beat at 25 beats per second. What is the period of the wing beating in seconds?

• A. 0.04 s
• B. 0.25 s
• C. 0.4 s
• D. 4 s

Correct answer: A

Rationale: The period represents the time for one complete cycle of the wing beating. To calculate the period, you take the reciprocal of the frequency. In this case, with the wings beating at 25 beats per second, the period is 1/25, which equals 0.04 seconds. Therefore, choice A, 0.04 seconds, is correct. Choices B, C, and D are incorrect because they do not reflect the correct calculation of the period based on the given frequency of 25 beats per second.

2. Two objects attract each other with a gravitational force of 12 units. If you double the mass of both objects, what is the new force of attraction between them?

• A. 3 units
• B. 6 units
• C. 24 units
• D. 48 units

Correct answer: C

Rationale: The gravitational force between two objects is directly proportional to the product of their masses. When you double the masses of both objects, the force of attraction between them increases by a factor of 2 x 2 = 4. Therefore, the new force of attraction between the two objects will be 12 units x 4 = 24 units. Choices A, B, and D are incorrect because doubling the mass results in a quadruple increase in force, not a linear one.

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. Certain non-Newtonian fluids exhibit shear thickening behavior. In this case, the fluid's viscosity:

• A. Remains constant with increasing shear rate
• B. Decreases with increasing shear rate (shear thinning)
• C. Increases with increasing shear rate
• D. Depends solely on the applied pressure

Correct answer: C

Rationale: When a non-Newtonian fluid exhibits shear thickening behavior, its viscosity increases with increasing shear rate. This means that as more force is applied to the fluid, its resistance to flow also increases, resulting in a higher viscosity. This phenomenon is opposite to shear thinning, where viscosity decreases with increasing shear rate. Therefore, in the case of shear thickening behavior, the correct answer is that the fluid's viscosity increases with increasing shear rate. Choices A, B, and D are incorrect because shear thickening behavior specifically involves an increase in viscosity with increasing shear rate, not remaining constant, decreasing, or depending on applied pressure.

5. What force was applied to the object that was moved if 100 Nâ‹…m of work is done over 20 m?

• A. 5 N
• B. 80 N
• C. 120 N
• D. 2,000 N

Correct answer: A

Rationale: Work is calculated using the formula Work = Force x Distance. Given that 100 Nâ‹…m of work is done over 20 m, we can rearrange the formula to solve for Force. Force = Work / Distance. Plugging in the values, we get Force = 100 Nâ‹…m / 20 m = 5 N. Therefore, the force applied to the object that was moved is 5 N. Choice B (80 N) is incorrect because it doesn't match the calculated force of 5 N. Choice C (120 N) is incorrect as it is higher than the calculated force. Choice D (2,000 N) is incorrect as it is significantly higher than the correct force of 5 N.

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