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Two objects attract each other with a gravitational force of 12 units. If the distance between them is halved, what is the new force of attraction between the two objects?
- A. 3 units
- B. 6 units
- C. 24 units
- D. 48 units
Correct Answer: C
Rationale: The gravitational force between two objects is inversely proportional to the square of the distance between them. When the distance is halved, the new force of attraction will be 12 units x (1/(0.5)^2) = 12 units x 4 = 24 units. Therefore, the correct answer is C. Choice A and B are incorrect as they do not consider the inverse square law of gravitational force. Choice D is incorrect as reducing the distance between the objects does not lead to a squared increase in force.
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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).
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.
A 25-cm spring stretches to 28 cm when a force of 12 N is applied. What would its length be if that force were doubled?
- A. 31 cm
- B. 40 cm
- C. 50 cm
- D. 56 cm
Correct Answer: A
Rationale: When the 12 N force stretches the spring from 25 cm to 28 cm, it causes a length increase of 28 cm - 25 cm = 3 cm. Therefore, each newton of applied force causes an extension of 3 cm / 12 N = 0.25 cm/N. If the force is doubled to 24 N, the spring would extend by 24 N 0.25 cm/N = 6 cm more than its original length of 25 cm. Thus, the new length of the spring would be 25 cm + 6 cm = 31 cm. Choice A, 31 cm, is the correct answer as calculated. Choices B, C, and D are incorrect as they do not consider the relationship between force and extension in the spring, leading to incorrect calculations of the new length.
What is the kinetic energy of a 500-kg wagon moving at 10 m/s?
- A. 50 J
- B. 250 J
- C. 2.5 10^4 J
- D. 5.0 10^5 J
Correct Answer: C
Rationale: The formula for calculating kinetic energy is KE = 0.5 mass velocity². Given the mass of the wagon is 500 kg and the velocity is 10 m/s, we can substitute these values into the formula: KE = 0.5 500 kg (10 m/s)² = 0.5 500 kg 100 m²/s² = 25,000 J or 2.5 10â´ J. Therefore, the kinetic energy of the 500-kg wagon moving at 10 m/s is 2.5 10â´ J. Choice A (50 J) is incorrect because it is too low; Choice B (250 J) is incorrect as it does not match the correct calculation; Choice D (5.0 10^5 J) is incorrect as it is too high. The correct answer is C (2.5 10^4 J).
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.