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Mariulka [41]
2 years ago
8

A 5. 0-c charge is 10 m from a small test charge. what is the magnitude of the force experienced by a 1. 0 nc charge placed at t

he location of the test charge?
Physics
1 answer:
sveta [45]2 years ago
4 0

The magnitude of the force experienced by a 1. 0 c charge placed at the location of the test charge is 4.5*10^8N.

To find the answer, we have to know more about the force experienced between two charges.

<h3>How to find the force experienced between two charges?</h3>
  • It is given that,

                Q_1=5C\\Q_2=1C\\r=10m.

  • We have to find the force experienced between two charges,

               F=\frac{kQ_1Q_2}{r^2}

where, k=8.99*10^9.

  • Thus, by substitution, the force will be,

          F=\frac{8.99*10^9*5*1}{10^2} =4.5*10^8N

Thus, we can conclude that, the magnitude of the force experienced by a 1. 0 c charge placed at the location of the test charge is 4.5*10^8N.

Learn more about the force here:

brainly.com/question/4086258

#SPJ4

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A car slows down at -5.00 m/s² until it comes to a stop after travelling 15.0 m. What was the initial speed of the car?
lapo4ka [179]
<h2>Answer: 12.24m/s</h2>

According to <u>kinematics</u> this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.

Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:

V_{f}^{2}-V_{o}^{2}=2.a.d   (1)

Where:

V_{f} is the Final Velocity of the car. We are told "the car comes to a stop after travelling", this means it is 0.

V_{o} is the Initial Velocity, the value we want to find

a is the constant acceleration of the car (the negative sign means the car is decelerating)

d is the distance traveled by the car

Now, let's substitute the known values in equation (1) and find V_{o}:

0-V_{o}^{2}=2(-5m/{s}^{2})(15m)    (2)

-V_{o}^{2}=-150{m}^{2}/{s}^{2}    (3)

Multiplying by -1 on both sides of the equation:

V_{o}^{2}=150{m}^{2}/{s}^{2}    (4)

V_{o}=\sqrt{150{m}^{2}/{s}^{2}}    (5)

Finally:

V_{o}=12.24m/s >>>This is the Initial velocity of the car

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Clarge%20%5Ctextbf%7BHelp%20with%20this%20question%20%7D%20" id="TexFormula1" title="\large
PIT_PIT [208]

<h3>Hi there !</h3><h2>Option A is correct </h2>

<h3> Please refer the attachment for explanation</h3><h2>Stay safe, stay healthy and blessed</h2><h2>Have a marvelous day</h2><h2>Thank you</h2>

8 0
2 years ago
If an object was accelerating at 10 m/s2, and a mass of 1 kg, what was size of the force acting on the object?
ryzh [129]

Answer:

10 N

Explanation:

f = ma

= 10m/s^2 * 1 kg

=10N

7 0
3 years ago
ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements.
vekshin1

Answer:

The answer is 500

Explanation:

6 0
3 years ago
According to Newton's<br> Ist law, what will an<br> object in motion tend<br> to do?
SCORPION-xisa [38]

Answer:

According to <em>Newton's first law of motion:</em>

<u>An object in motion tends to remain in motion unless an external force acts upon it.</u>

<u>It stays in motion with the same speed and goes in the same direction.</u>

<u></u>

<em>Hope this helped </em>

<em>:)</em>

4 0
3 years ago
Read 2 more answers
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