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Viktor [21]
3 years ago
8

The ball is dropped from a certain height and is falling to the ground. If its acceleration is a constant 9.8 meters per second

squared, how many seconds would it have dropped when the velocity is 49.0 m/s ????
A. 4.9 secs
B. 4 secs
C. 5 secs
D. 9.8 secs

Physics
1 answer:
lord [1]3 years ago
7 0
5 secs I did 49 divided by 9.8 and got 5
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Answer:

c) 4.2*10^{-5}C

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

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F=\frac{kq^{2}}{r^{2}}

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q=\sqrt{\frac{Fr^{2}}{k}}

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q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}

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