Explanation:
What exactly are u looking for?
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)
Answer:
Explanation:
This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:
which, in words, is
The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:
Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).
0 = 90.0v - 2.0 and
2.0 = 90.0v so
v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.
12 = .022t and
t = 550 seconds, which is the same thing as 9.2 minutes
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According to the law of conservation of energy, the potential energy is converted to kinetic energy. Remember, potential energy is calculated using height and weight. If the ball is on the ground, height is 0.