His hands being on the pole creates the thermal energy because of the friction being made by sliding down the pole.
Let's say the distance is D. Then the time going is D/10 sec. The time returning is D/20 s. The total time is 3D/20 s, and the total distance is 2D. The average speed for the round trip is (total distance)/(total time). That's (2D) ÷ (3D/20). That's (40D/3D) which is 13-1/3 m/s. (I thought it was going to depend on the distance, but it doesn't.)
Answer:
The speed of the particle is 2.86 m/s
Explanation:
Given;
radius of the circular path, r = 2.0 m
tangential acceleration,
= 4.4 m/s²
total magnitude of the acceleration, a = 6.0 m/s²
Total acceleration is the vector sum of tangential acceleration and radial acceleration

where;
is the radial acceleration

The radial acceleration relates to speed of particle in the following equations;

where;
v is the speed of the particle

Therefore, the speed of the particle is 2.86 m/s
Answer:
a) E = V/L x^
b) R = ρL/A = ρL/π(d/2)^2 = 4ρL/πd^2
c) I = V/R x^ = V/(4ρL/πd^2) x^ = πd^2*V/4ρL x^
d) J = I/A = [πd^2*V/4ρL x^]/π(d/2)^2 = V/ρL x^
e) ρJ = ρ(V/ρL x^) = V/L x^ = E