<span>At an instant when the displacement is equal to a/2,
Potential energy U = 1/2ka(square) where a is displacement.
when a= a/2
U = 1/4ka(square)
U = E/4
Potential Energy = 1/4 Total energy</span>
sorry its quite messy haha
Explanation:
It is given that,
Diameter of the circular loop, d = 1.5 cm
Radius of the circular loop, r = 0.0075 m
Magnetic field, 
(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :
I = 32.22 A
(b) The magnetic field on a current carrying wire is given by :
r = 0.00238 m
Hence, this is the required solution.
Answer:
Maximum Tension=224N
Minimum tension= 64N
Explanation:
Given
mass =8 kg
constant speed = 6m/s .
g=10m/s^2
Maximum Tension= [(mv^2/ r) + (mg)]
Minimum tension= [(mv^2/ r) - (mg)]
Then substitute the values,
Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N
Minimum tension= [8 × 6^2)/2 -(8×9.8)]
=64N
Hence, Minimum tension and maximum Tension are =64N and 2224N respectively
