Answer:
Given that
For A weight Wt= 22.7 N
m₁ = 22.7/10 = 2.27 kg
Force alone inclined plane
Wt₁ = m₁ g sin θ
Wt₁ =22.7 sin 17.2°
Wt₁ = 6.7 N
For B weight Wt₂= 34.5-N
m₂ = 3.45 kg
coefficient of friction ,μ= 0.219
θ = 17.2 degree
The friction force on the block A
Fr= μ m₁ g cos θ
Fr= 0.216 x 22.7 x cos 17.2°
Fr= 4.68 N
Lets take acceleration of system is a m/s²
Tension = T
From Newtons law
Wt₂ - Wt₁ - Fr = (m₁ +m₂) a
34.5 - 6.7 - 4.68 = (2.27 + 3.45 ) a
a= 4.05 m/s²
Block B
Wt₂ - T = m₂ a
T = 34.5 - 4.05 x 3.45
T= 20.52 N
Answer:
Elastic potential energy,
Explanation:
Charge,
Potential, V = 50 V
It is required to find the electric potential energy in a capacitor stored in it. The formula of the electric potential energy in a capacitor is given by :
So, the electric potential energy stored in the capacitor is