Dissolves limestone and other rocks.
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:
The cross product of two vectors is given by

Where, θ be the angle between the two vectors and \widehat{n} be the unit vector along the direction of cross product of two vectors.
Here, K x i = - j
As K is the unit vector along Z axis, i is the unit vector along X axis and j be the unit vector along axis.
The direction of cross product of two vectors is given by the right hand palm rule.
So, k x i = j
j x i = - k
- j x k = - i
i x i = 0
Explanation:
We have,
Length of a metal rod is 55 cm or 0.55 m
Change in length is 0.2 cm or 0.002 m
It is required to find the change in temperature of a metal rod. The coefficient of linear expansion is given by :

is the change in temperature

So, the change in temperature is 303.03 degrees Celsius.