Answer:
Explanation:
Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J
Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J
But ,
4 * 1875000 = 7500000
so the KE has increased by 4 times.
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
.
The coal is urned to heat up water. this produces steam. the steam turns a turbine that turns a generator which provided energy yhat can be transferred into electrisity
Answer:
-2040 m/s²
Explanation:
Taking toward the wall to be positive, the initial velocity is 10.1 m/s and the final velocity is -8.3426 m/s.
Average acceleration is the change in velocity over change in time.
a = Δv / Δt
a = (-8.3426 m/s − 10.1 m/s) / 0.00905 s
a = -2040 m/s²