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.
The correct choice is
B.
Particles at the bottom of the water carry heat energy to the top of the water.
when pot of water is heater, the bottom of pot gets heated. the particles of water in contact with the bottom of the pot gets heat through conduction. after getting heat, these particles of water near the bottom, move away towards top and their position is taken by cooler particles from top. that way heat travels
Answer:
Explanation:
<u>Vertical Launch Upwards</u>
In a vertical launch upwards, an object is launched vertically up without taking into consideration friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:
The tennis ball was thrown straight up with a speed of v0=22.5 m/s. The acceleration of gravity is g=9.81\ m/s^2, thus:
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
The object is moving, so at different times, it has different displacement. I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.
Displacement is the distance and the direction FROM (the position at the beginning) TO (the position at the end).
At the beginning ... time=0 ... the position is 1 meter.
At the end ... time=5 ... the position is zero.
The distance FROM the beginning TO the end is (zero - 1m) . That's <em>-1m </em>.
