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alexgriva [62]
2 years ago
8

A car was moving a 14 m/s. After 30 seconds, it’s speed increased to 20 m/s. What was its acceleration during this time?

Physics
2 answers:
Anuta_ua [19.1K]2 years ago
5 0

Answer:

acceleration = 0.2 m/s/s

Explanation:

initial velocity u = 14

final velocity v = 20

time = 30

acceleration = ?

v = u + at

20 = 14 + 30a

30a = 6

a = 0.2 m/s/s

Alexeev081 [22]2 years ago
4 0

Answer:

For the aceleration we have:

V = Vi + a * t

Being:

V = 20 m/s

Vi = 14 m/s

a = ?

t = 30 s

Then, lets replace acording the formula:

20 m/s = 14 m/s + a * 30 s

Clear "a":

a = (20 m/s - 14 m/s) / 30 s

a = (6 m/s) / 30 s

a = 0,2 m/s²

The aceleration of the car is <u>0,2 meters per second squared.</u>

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What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
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The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 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 3.00 \times 10^{4} V/m?

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