Answer:
a motor is used to covert electrical energy to mechanical energy
Answer:
Option B. 6.25 J/S
Explanation:
Data obtained from the question include:
t (time) = 2secs
F (force) = 50N
d (distance) = 0.25m
P (power) =?
The power can be obtained by using the formula P = workdone/time.
P = workdone / time
P = (50 x 0.25)/ 2
P = 6.25J/s
Answer:
V = -RC (dV/dt)
Solving the differential equation,
V(t) = V₀ e⁻ᵏᵗ
where k = RC
Explanation:
V(t) = I(t) × R
The Current through the capacitor is given as the time rate of change of charge on the capacitor.
I(t) = -dQ/dt
But, the charge on a capacitor is given as
Q = CV
(dQ/dt) = (d/dt) (CV)
Since C is constant,
(dQ/dt) = (CdV/dt)
V(t) = I(t) × R
V(t) = -(CdV/dt) × R
V = -RC (dV/dt)
(dV/dt) = -(RC/V)
(dV/V) = -RC dt
∫ (dV/V) = ∫ -RC dt
Let k = RC
∫ (dV/V) = ∫ -k dt
Integrating the the left hand side from V₀ (the initial voltage of the capacitor) to V (the voltage of the resistor at any time) and the right hand side from 0 to t.
In V - In V₀ = -kt
In(V/V₀) = - kt
(V/V₀) = e⁻ᵏᵗ
V = V₀ e⁻ᵏᵗ
V(t) = V₀ e⁻ᵏᵗ
Hope this Helps!!!
