Starting from the angular velocity, we can calculate the tangential velocity of the stone:
Then we can calculate the angular momentum of the stone about the center of the circle, given by
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.
Substituting the data of the problem, we find
The solution for this problem would be:(10 - 500x) / (5 - x)
so start by doing:
x(5*50*2) - xV + 5V = 0.02(5*50*2)
500x - xV + 5V = 10
V(5 - x) = 10 - 500x
V = (10 - 500x) / (5 - x)
(V stands for the volume, but leaves us with the expression for x)
Answer:
Explanation:
Given that
Length= 2L
Linear charge density=λ
Distance= d
K=1/(4πε)
The electric field at point P
So
Now by integrating above equation
= Heat released to cold reservoir
= Heat released to hot reservoir
= maximum amount of work
= temperature of cold reservoir
= temperature of hot reservoir
we know that
eq-1
maximum work is given as
= -
using eq-1
= 
-
Answer: The answer to the question is 0.25 ohms
Explanation:
R = u x/A .......1
where u is chosen as the resistivity of
the rod, A is area of the rod and x is
chosen as the length of the rod.
Let R* be the resistance of the
lengths of the rod across the adjoints
Then R* = u1/4.......2
Comparing equation 1 and 2
R* = 1/4
=0.25ohms
