Nokea
The Prandtl number (Pr) is a dimensionless property relating:
- A. Viscosity and thermal diffusivity
- B. Density and pressure
- C. Surface tension and pressure
- D. Reynolds number and flow regime
Correct Answer: A
Rationale: The Prandtl number (Pr) is a dimensionless number used to characterize fluid flow. It is the ratio of momentum diffusivity to thermal diffusivity. In simpler terms, it relates the ability of a fluid to conduct heat to its ability to conduct momentum. Therefore, the correct relationship is between viscosity and thermal diffusivity, making choice A the correct answer. Choices B, C, and D are incorrect because they do not represent the properties that the Prandtl number relates.
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Marilyn is driving to a wedding. She drives 4 miles south before realizing that she left the gift at home. She makes a U-turn, returns home to pick up the gift, and sets out again driving south. This time, she drives 1 mile out of her way to pick up a friend. From there, they continue 5 miles more to the wedding. Which of these statements is true about Marilyn's trip?
- A. The displacement of her trip is 6 miles, and the distance traveled is 6 miles.
- B. The displacement of her trip is 14 miles, and the distance traveled is 14 miles.
- C. The displacement of her trip is 8 miles, and the distance traveled is 14 miles.
- D. The displacement of her trip is 6 miles, and the distance traveled is 14 miles.
Correct Answer: C
Rationale: Marilyn's displacement is calculated based on her final position relative to the starting point. She drives 1 mile to pick up her friend, then 5 miles more to the wedding, totaling 6 miles after returning to her home. So, the correct displacement is 8 miles south from her starting point (4 miles to the gift + 4 miles return + 1 mile to the friend + 5 miles to the wedding). The total distance traveled is 14 miles (adding all the distances). Choice A is incorrect because it miscalculates the displacement. Choice B is incorrect as it overestimates both the displacement and distance traveled. Choice D is incorrect as it underestimates the displacement.
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).
An object with a charge of 3 μC is placed 30 cm from another object with a charge of 2 μC. What is the magnitude of the resulting force between the objects?
- A. 0.6 N
- B. 0.18 N
- C. 180 N
- D. 9 10−12 N
Correct Answer: B
Rationale: To find the magnitude of the resulting force between two charges, we use Coulomb's Law:
F = k (|q1 q2|) / r²
Where:
F is the force
k is Coulomb's constant (8.99 10â¹ N·m²/C²)
q1 and q2 are the charges
r is the distance between the charges
Plugging in the values:
F = (8.99 10â¹) (3 10â»â¶) (2 10â»â¶) / (0.3)² = 0.18 N.
Therefore, the magnitude of the resulting force is 0.18 N.
An object has a constant velocity of 50 m/s and travels for 10 s. What is the acceleration of the object?
- A. 0 m/s²
- B. 5 m/s²
- C. 60 m/s²
- D. 500 m/s²
Correct Answer: A
Rationale: The acceleration of an object is defined as the rate of change of its velocity. When an object has a constant velocity, it means there is no change in its speed or direction. In this case, the object maintains a constant velocity of 50 m/s for 10 seconds, which implies that there is no change in velocity. Therefore, the acceleration of the object is 0 m/s² as there is no acceleration or deceleration happening. Choices B, C, and D are incorrect because acceleration is the change in velocity over time, and in this scenario of constant velocity, the acceleration is 0 m/s².
The specific heat capacity of tin is 217 J/(g°C). Which of these materials would require about twice as much heat as tin to increase the temperature of a sample by 1°C?
- A. Copper [0.3844 J/(g°C)]
- B. Iron [0.449 J/(g°C)]
- C. Gold [0.1291 J/(g°C)]
- D. Aluminum [0.904 J/(g°C)]
Correct Answer: D
Rationale: The correct answer is D: Aluminum. The specific heat capacity of aluminum is 0.904 J/(g°C), which is approximately 4 times that of tin. For a material to require about twice as much heat as tin to increase the temperature by 1°C, it should have a specific heat capacity roughly double that of tin. Therefore, aluminum fits this criterion better than the other options. Gold has a much lower specific heat capacity than tin, so it would require less, not more, heat to increase the temperature by 1°C. Copper and Iron also have specific heat capacities lower than tin, making them incorrect choices for requiring twice as much heat as tin.