A ball is thrown vertically upwards at 19.6 m/s. For its complete trip (up and back down to the starting position), its average
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:
t = 13.7 s or t = 14 s with proper significant figures
Explanation:
The initial speed is 0 m/s since the car starts from rest, acceleration is 5.5 m/s2 and distance is 523 m.
Since we have initial speed, acceleration and distance we can use the following formula to find the time. We can now use algebra to work out our answer.
d = vt + at²
523 = (0)t + ()(5.5)t²
523 = 2.8t²
186.8 = t²
13.7 s = t
(t = 14 s with proper significant figures)