Before the student releases the cart by cutting the tinsel string, what forces are acting on the cart?
1 answer:
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The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q =
Therefore, substitute the values into the above formula as follows.
Q =
=
=
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .
Answer:
Hz
Explanation:
In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;
V(t) = V sin (ωt + Ф) -----------------(i)
Where;
V = amplitude value of the voltage
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
Ф = phase difference between voltage and current.
<u><em>Now,</em></u>
From the question,
V(t) = 230 sin (100t) ---------------(ii)
<em><u>By comparing equations (i) and (ii) the following holds;</u></em>
V = 230
ω = 100
Ф = 0
<em><u>But;</u></em>
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f = Hz
Therefore, the frequency of the voltage is Hz