Answer:
0.0025H
Explanation:
I didn't come here to be part of this all I wanted is just information for my research
Answer:
Torque decreases .
Explanation:
The tape is pulled at constant speed , speed v is constant , so there is
v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .
Let it be α . Let I be moment of inertia of reel .
Reel is in the form of disc
I = 1/2 m r²
I x α = torque
1/2 m r² x α = torque
As the reel is untapped , its mass decreases , r also decreases , so torques also decreases .
Answer:

Explanation:
From the Question we are told that:
Mass 
Coefficient of kinetic friction 
Generally the equation for Frictional force is mathematically given by



Generally the Newton's equation for Acceleration due to Friction force is mathematically given by



Therefore



The answer for this change in the magnitude of momentum is the same for both because momentum is always conserved so both vehicles have the identical change.
So for determining who has the greater change in kinetic energy, momentum (P) = mv so P^2 = m^2 v^2 P^2 / 2m = 1/2 m v^2 = energy So the weightier the mass the smaller the energy change for the same momentum change so in here, the car has a greater change in kinetic energy.
These are called neap tides!
Hope this helps :)