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vovikov84 [41]
3 years ago
15

What is the magnitude of the force F on the 1.0 nC charge in the figure ?

Physics
1 answer:
Dmitriy789 [7]3 years ago
3 0
In this problem, the 2.0 nC charges both act forces of equal magnitude on the 1.0 nC charge. Since there x-components are equal but opposite, they cancel out and only the y-components remain and add up. This can be used then to calculate for the magnitude of the total electrostatic force.

F = 2 * q1 * q2 / 4 / pi / eps_0 / r^2 * sqrt(3) / 2
F = 2 * 1.0e-9 * 2.0e-9 / 4 / pi / 8.854e-12 / (1.0e-2)^2 * sqrt(3) / 2
F = 3.11e-4 N

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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The Hubble Space Telescope is in a "Low-Earth Orbit" with an orbital period of 95min. Calculate the altitude of the HST's orbit
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Answer:I don’t know please tell me

Explanation:

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An electric fan is running on HIGH. After the LOW button is pressed, the angular speed of the fan decreases to 84.7 rad/s in 2.9
ale4655 [162]

The initial angular speed of the fan will be 55.0 rad/sec. The angular speed of the fan decreases to 84.7 rad/s in 2.96 s.

<h3>What is angular acceleration?</h3>

Angular acceleration is defined as the pace of change of angular velocity with reference to time.

Given data;

Final angular speed,\rm \omega_f = 84.7 rad/s

Initial angular speed, \rm \omega_i = ?

Time period,t= 2.96 s

Angular deceleration = 47.2 rad/s²

\rm \alpha =\frac{\omega_f-\omega_i}{t} \\\\\rm 47.2 =\frac{84.7-\omega_i}{2.96} \\\\ \omega_i = 55.0  \ rad/sec

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7 0
2 years ago
12. An unbalanced 6.0 newton force acts eastward on an object for 3.0 seconds. The impulse
meriva

Answer:

Impulse=18Ns

Explanation:

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3 0
3 years ago
Acceleration occurs whenever an ___________________ force acts on an object.
leonid [27]
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8 0
4 years ago
Read 2 more answers
A typical running track is an oval with 74-m-diameter half circles at each end. A runner going once around the track covers a di
nirvana33 [79]

Answer:

The acceleration towards the center will be 0.43\ m/s^2

Explanation:

Given the running track is an oval shape, and the diameter of each half-circle is 74 meters.

Also, the runner took 1 minute and 40 seconds to complete 400 m one round of the track.

We need to find the acceleration towards the center.

First, we will find the speed.

v=\frac{d}{t}

Where v is the speed.

d is the distance covered by the rider that is 400 meters.

t is the time taken by the rider to complete the lap which is 1 minute and 40 seconds.  (60\ s +40\ s) = 100 seconds.

So,

v=\frac{400}{100}=4\ m/s

And

a_c=\frac{v^2}{r}

Where a_c is the acceleration towards the center.

r is the radius which will be the half of the diameter 74 meters.

Hence, the radius will be 37 meters.

a_c=\frac{4^2}{37} \\a_c=\frac{16}{37}=0.43\ m/s^2

So, the centripetal acceleration of the rider will be 0.43\ m/s^2

4 0
3 years ago
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