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For a compressible fluid subjected to rapid pressure changes, sound wave propagation becomes important. The speed of sound (c) depends on the fluid's:
- A. Density (Ï) only
- B. Viscosity (μ) only
- C. Density (Ï) and Bulk modulus
- D. Density (Ï) and Surface tension (γ)
Correct Answer: C
Rationale: In a compressible fluid, the speed of sound (c) depends on both the fluid's density (Ï) and Bulk modulus. Density affects the compressibility of the fluid, while Bulk modulus represents the fluid's resistance to compression and plays a crucial role in determining the speed of sound in a compressible medium. Viscosity and surface tension do not directly impact the speed of sound in a compressible fluid subjected to rapid pressure changes. Therefore, the correct answer is C.
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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.
A closed system undergoes a cyclic process, returning to its initial state. What can be said about the net work done (Wnet) by the system over the entire cycle?
- A. Wnet is always positive.
- B. Wnet is always negative.
- C. Wnet can be positive, negative, or zero.
- D. Wnet is equal to the total heat transferred into the system (dQ ≠0 for a cycle).
Correct Answer: C
Rationale: For a closed system undergoing a cyclic process and returning to its initial state, the net work done (Wnet) over the entire cycle can be positive, negative, or zero. This is because the work done is determined by the area enclosed by the cycle on a P-V diagram, and this area can be above, below, or intersecting the zero work axis, leading to positive, negative, or zero net work done. Choice A is incorrect because Wnet is not always positive; it depends on the specific path taken on the P-V diagram. Choice B is incorrect as Wnet is not always negative; it varies based on the enclosed area. Choice D is incorrect because Wnet is not necessarily equal to the total heat transferred into the system; it depends on the specifics of the cycle and is not a direct relationship.
A 110-volt hair dryer delivers 1,525 watts of power. How many amperes does it draw?
- A. 167.75 amperes
- B. 1.635 amperes
- C. 1.415 amperes
- D. 13.9 amperes
Correct Answer: D
Rationale: To determine the amperes drawn by the hair dryer, we use the formula: Amperes = Watts / Volts. The hair dryer operates at 1,525 watts with 110 volts. Dividing 1,525 watts by 110 volts yields 13.9 amperes. Therefore, the correct answer is 13.9 amperes. Choices A, B, and C are incorrect because they do not result from the correct calculation using the formula.
Which of the following describes a vector quantity?
- A. 5 miles per hour due southwest
- B. 5 miles per hour
- C. 5 miles
- D. None of the above
Correct Answer: A
Rationale: A vector quantity is characterized by both magnitude and direction. In the provided options, choice A, '5 miles per hour due southwest,' fits this definition as it includes both the magnitude (5 miles per hour) and the direction (southwest), making it a vector quantity. Choices B and C only provide the magnitude without indicating any direction, hence they do not represent vector quantities.
A 5-cm candle is placed 20 cm away from a concave mirror with a focal length of 10 cm. What is the image distance of the candle?
- A. 20 cm
- B. 40 cm
- C. 60 cm
- D. 75 cm
Correct Answer: C
Rationale: To find the image distance of the candle, we use the mirror formula: 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. In this case, the focal length f = 10 cm and the object distance do = 20 cm. Substituting these values into the formula gives us 1/10 = 1/20 + 1/di. Solving for di, we get di = 60 cm. Therefore, the image distance of the candle is 60 cm. Choice A (20 cm) is incorrect because it represents the object distance, not the image distance. Choice B (40 cm) is incorrect as it does not consider the mirror formula calculation. Choice D (75 cm) is incorrect as it does not match the correct calculation based on the mirror formula.