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.
Answer:
ΔV=0.484mV
Explanation:
The potential difference across the end of conductor that obeys Ohms law:
ΔV=IR
Where I is current
R is resistance
The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A
R=(pL)/A
Given data
Length L=3.87 cm =0.0387m
Diameter d=2.11 cm =0.0211 m
Current I=165 A
Resistivity of aluminum p=2.65×10⁻⁸ ohms
So
ΔV=IR

ΔV=0.484mV
Answer:
No. Your friend will not hear the clap when he/she sees it.
Explanation:
It takes time for sound waves to go through a large area. It takes longer if people are in the area rather than it is empty.
To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

Through the aforementioned formula we will have to

The particulate part of the rest, so the final speed would be



Now from Newton's second law we know that

Here,
m = mass
a = acceleration, which can also be written as a function of velocity and time, then

Replacing we have that,


Therefore the force that the water exert on the man is 1386.62