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:
Explanation:
In all atoms, the number of protons = number of electrons, as a result the atom is neutral. Losing or gaining electrons will make the atom electrically charged and we call an electrically charged atom an ion.
Ca 2+ would be the symbol because losing two negative electrons makes calcium's nucleus more positive by two protons.
Answer:
Cp= 0.44 J/g.C
This is heat capacity of metal.
Explanation:
From energy conservation
Heat lost by metal = Heat gain by water +Heat gain by calorimeter
Because here temperature of metal is high that is why it loose the heat.The temperature of water and calorimeter is low that is why they gain the heat.
final temperature is T= 30.5 C
We know that sensible heat transfer given as
Q= m Cp ΔT
m=Mass
Cp=Specific heat capacity
ΔT=Temperature difference
By putting the values
55 x Cp ( 99.5 - 30.5) = 40 x 4.184 ( 30.5- 21 ) + 10 x ( 30.5 - 21)
Cp ( 99 .5- 30.5) = 30.65
Cp= 0.44 J/g.C
This is heat capacity of metal.
Centripetal acceleration points from the object toward the center of the circular path it's traveling.
Answer:
D; The microscope and magnifying glass block out the light, which allows the naked eye to focus on the object.
Explanation:
This is to prevent chromatic abberation
