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.
Alpha particles travel through the air they collide with oxygen and nitrogen molecules. While they collide with these molecules, they lose some energy until all energy are used up and they are absorbed. These particles can be absorbed by a sheet of paper or by the air. On the other hand, beta particles and gamma particles move faster than the alpha particles and are poor at ionizing atoms or molecules thus it takes more of the material to be able to absorb these particles.
Answer:
Ax = 0
Ay = 6 m
Bx = 8 cos phi = cos 34 = 6.63 m
By = 8 sin phi = 8 sin (-34) = -4.47 m
Rx = Ax + Bx = 0 + 6.63 = 6.63 m
Ry = Ay + By = 6 - 4.47 = 1.53 m
R = (6.63^2 + 1.53^2)^1/2 = 6.80 m
tan theta = Ry / Rx = 1.53 / 6.8 = ,225
theta = 12.7 deg
Answer:
a) 0.138J
b) 3.58m/S
c) (1.52J)(I)
Explanation:
a) to find the increase in the translational kinetic energy you can use the relation
where Wp is the work done by the person and Wg is the work done by the gravitational force
By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:
the change in the translational kinetic energy is 0.138J
b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:
where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:
the new speed is 3.58m/s
c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy
hence, the change in Er is about 1.52J times the initial rotational energy
