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 kinetic energy of the phone right before it hits the ground is 9J.
<h3>
Kinetic energy of the phone</h3>
The kinetic energy of the phone right before it hits the ground is calculated as follows;
K.E = ¹/₂mv²
where;
- m is mass of the phone
- v is velocity of the phone
K.E = ¹/₂(0.08)(15)²
K.E = 9 J
Thus, the kinetic energy of the phone right before it hits the ground is 9J.
Learn more about kinetic energy here: brainly.com/question/25959744
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Answer: if it is in inches than it is 21.844 cm
Explanation:
Answer:
Explanation:
A right triangle is formed, in which the vertical elevation is the opposite cathetus and the horizontal distance is the adjacent cathetus, since we know these two values, we can calculate the angle of inclination using the definition of tangent:
Am sorry what can you be more specific