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 .
<span>The correct answer is blue. If you look at a luminosity star chart, called the Hertzsprung Russell Diagram, you will see the measurement of luminosity on the left side, and you will see a curve of stars with different colors (which correlate to the colors of the stars). Look for 30 on the luminosity measurement (look between 1 and 100). Then, move horizontally across the diagram until you hit the stars, whose color will be blue. Thus, blue is the answer.</span>
Answer:
v= 335 m/s
2∆t= 0.75 s
∆x= v.∆t → ∆x= 335×½×0.75 = 125.625 m
The answer will be 50N.
This is because the spring reads weight and weight is mass times acceleration due to gravity.5kg*10m/s2=50N
Mount. everest is 5.499 miles
