Answer:
V = 280.15 V
Explanation:
" The complete question is attached with figure"
Given:
- The capacitance of the capacitor C = 10 nF
- The amount of mass attached to motor m = 4 grams
- The amount of distance it is to be lifted h = 1 cm
- Ignore all other losses in the system
Find:
- The voltage required to lift the mass m through distance h?
Solution:
- The conservation of energy for the entire system is written as:
Work_gravity = U_c
Where,
Work_gravity: Work done by gravity on mass m
U_c: The amount of energy stored in a capacitor
m*g*h = 0.5*C*V^2
V^2 = 2*m*g*h / C
V = sqrt ( 2*m*g*h / C )
Plug in the values:
V = sqrt ( 2*0.004*9.81*0.01 / 10*10^-9 )
V = sqrt ( 78,480)
V = 28.15 V
In this exercise we have to use the capacitor knowledge to calculate the system losses, this is:
V = 280.15 V
From the information given in the statement we find that:
Remembering that in a closed system we have the conservation of energy as:
[tex]Work_g = U_c[/tex]
substituting the known values into:
[tex]m*g*h = 0.5*C*V^2\\V^2 = 2*m*g*h / C\\ V = sqrt ( 2*m*g*h / C )\\V = sqrt ( 2*0.004*9.81*0.01 / 10*10^{-9} )\\ V = sqrt ( 78,480)\\ V = 28.15 V[/tex]
See more about capacitor at brainly.com/question/14048432
