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:
See the explanation below,
Explanation:
Momentum is defined as the product of mass by Velocity. In this way, we will replace the following equation in each of the cases.
where:
P = momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
A)
B)
![P=100*5\\P=500[kg*m/s]](https://tex.z-dn.net/?f=P%3D100%2A5%5C%5CP%3D500%5Bkg%2Am%2Fs%5D)
C)
![P=1200*15\\P=18000[kg*m/s]](https://tex.z-dn.net/?f=P%3D1200%2A15%5C%5CP%3D18000%5Bkg%2Am%2Fs%5D)
D)
![P=15*1000\\P=15000[kg*m/s]](https://tex.z-dn.net/?f=P%3D15%2A1000%5C%5CP%3D15000%5Bkg%2Am%2Fs%5D)
From higher to lower momentum
C,D,B,A
Answer:
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sana maka tulong good luck (◍•ᴗ•◍)❤
Given:
u = 0, initial velocity
v = 125 m/s, final velocity
s = 0.0800 m, distance traveled
9.20 g, the mass of the pellet
If the acceleration is a, then
0² + 2*(a m/s²)*(0.800 m) = (125 m/s)²
1.6a = 15625
a = 9765.625 m/s²
Calculate the force.
F = (9.20 x 10⁻³ kg)*(9765.625 m/s²) = 98.84 N
Answer: 98.8 N (nearest tenth)
