Answer:
Explanation:
initial velocity u = 32.7 m /s
final velocity v = 50.3 m /s
displacement s = 44500 m
acceleration a = ?
v² = u² + 2 a s
50.3² = 32.7² + 2 x a x 44500
2530.09 = 1069.29 + 89000a
a .016 m /s²
time taken t = ?
v = u + at
50.3 = 32.7 + .016 t
t = 1100 s
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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
brainly.com/question/14883923
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.
Cadences.
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Acceleration=(speed end - speed start)/ time
Data:
speed end=4 m/s
speed start=0 m/s
time=2.5 s
acceleration=(4 m/s - 0 m/s)/2.5 s=1.6 m/s²
Answer: the acceleration would be 1.6 m/s²
