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tia_tia [17]
3 years ago
6

What is the electric force acting between two charges of -0.0050 C and

Physics
1 answer:
love history [14]3 years ago
6 0

The electric force is -3.6\cdot 10^8 N (attractive)

Explanation:

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

F=k\frac{q_1 q_2}{r^2}

where:

k=9\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, we have the following:

q_1 = -0.0050 C (charge 1)

q_2 = +0.0050 C (charge 2)

r = 0.025 m (distance)

Substituting, we find the electric force between the two charges:

F=(9\cdot 10^9) \frac{(-0.0050)(0.0050)}{(0.025)^2}=-3.6\cdot 10^8 N

And the negative sign means the force is attractive, since the two charges have opposite sign.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if
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Answer:

a) If the mass is doubled, then the period is increased by \sqrt{2}. Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

We have a block-spring system, whose angular frequency (\omega) is defined by the following formula:

\omega = \sqrt{\frac{k}{m} } (1)

Where:

k - Spring constant, measured in newtons per meter.

m - Mass, measured in kilograms.

And the period (T), measured in seconds, is determined by the following expression:

T = \frac{2\pi}{\omega} (2)

By applying (1) in (2), we get the following formula:

T = 2\pi\cdot \sqrt{\frac{m}{k} }

a) If the mass is doubled, then the period is increased by \sqrt{2}. Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

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Answer:

3.5m/s^2

Explanation:

From Newton's second Law of Motion

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Where F is the applied force, m is the mass of the object and a is the acceleration.

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