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The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
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Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.
Answer:
Approximately 
(assuming that the acceleration due to gravity is
.)
Explanation:
Assuming that the weight on this 72-kg skydiver would be 
(points downwards.)
Air resistance is supposed to act in the opposite direction of the motion. Since this skydiver is moving downwards, the air resistance on the skydiver would point upwards.
Therefore, the net force on this skydiver should be the difference between the weight and the air resistance on the skydiver:
.
Apply Newton's Second Law of motion to find the acceleration of this skydiver:
.
Answer:
=3.5 m/s
Explanation:
y = x tanθ - 1/2 g x² / (u²cos²θ )
y = 0.25 , x = 0.5, θ = 40°
.25 = .50 tan40 - .5 x 9.8x x²/ u²cos²40
.25 = .42 - 2.0875/u²
u = 3.5 m / s.
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