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
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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:
B.
Explanation:
According to CDC, if you’re doing moderate-intensity activity, you can talk but not sing during the activity.
Beat frequency is given by the difference of two frequencies played together

given that


Now


Answer:
The motion of an object is accelerated when its speed increases.
Answer:
a) True. The image of the mite is virtual
e) True. The image must be within the focal length of the eyepiece len
Explanation:
Let's review the general characteristics of compound microscopes
Formed by two converging lenses
Magnification is
M = -L/fo 0.25/fe
Where fo is the focal length of the objective lens and fe is the focal length of the ocular lens, L is the tube length
Let's review the claims
a) True. The image of the mite is virtual
b) False. The effect is the opposite of the magnification equation
c) False. The objective lens forms a real image
d) False. As the seal distance increases the magnification decreases
e) True. The image must be within the focal length of the eyepiece len