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:
I go with D but what you can do is u can read each and choose the one that sounds right Hope this helps
Answer:
v = 44,16 m/s
Explanation:
We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:
Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:
Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.
v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s 
So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.
I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D
The answer would be c.
your equation should look like:
reactants-----> products
