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Liono4ka [1.6K]
3 years ago
14

A spring stretches by 0.014 m when a 3.7-kg object is suspended from its end. How much mass should be attached to this spring so

that its frequency of vibration is f

Physics
1 answer:
lidiya [134]3 years ago
6 0

Explanation:

Below is an attachment containing the solution.

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two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t
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The charges have opposite sign and magnitude 6 \mu C

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

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where:

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q_1, q_2 are the two charges

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In this problem, we have:

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The two charges have same magnitude, so

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So we can rewrite the equation as

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

And solving for q:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

q_1 = 6 \mu C\\q_2 = -6 \mu C

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

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