Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J
Explanation:
Given that;
Mass M1 = 7.0 kg
r = 3.0/2 m = 1.5 m
Mass M2 = 21 kg
we know that G = 6.67 × 10⁻¹¹ N.m²/kg²
work done by an external agent W = -2GM2M1 / r
so we substitute
W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5
W = -1.96098 × 10⁻⁸ / 1.5
W = -1.3 × 10⁻⁸ J
Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J
True is the correct answer
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.
The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-
V∝ R [ where v is the rate of diffusion and r is the ratio of surface area to volume]
As per the question,the ratio of surface area to volume for a sphere is given 
The surface area to volume ratio for right circular cylinder is given 
Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.
Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.
