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:
Explanation:
Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:
Where Vf is the final velocity of the object, (in our case 80 m/s)
Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)
and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.
Therefore we have:
Notice that the units of acceleration in the SI system are 
(meters divided square seconds)
Answer:
real, inverted, and smaller than the object
Explanation:
When the object is placed beyond the center of curvature, the image will formed between the focus and the center of curvature. The size of the image is diminished and its nature is real and inverted.
The whole description is shown in the attached figure. It is clear that the size of the image is smaller than the object.
Answer:
714.285s
Explanation:
use relative velocity
8-4.5 = 3.5m/s
x = 2500m
2500/3.5 = 714.285s = 700s (with sig figs)
13-16 that is where they’re located at
