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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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Answer:

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   3. Methane, Ammonia, Carbon Dioxide.

Explanation:

Protoplanetry disk is the disk of gases and clouds of dust that rotates around the newly made star. The temperature of the protoplanetry disk actually determines the type of the planet that is to be formed. Inner part of the protoplanetry disk is closer to the sun thats why it is the hottest and denser part and composed of the materials like Iron, silicates, carbon as they have high melting points. Then comes those materials that exist in the solid form at lower temperatures such as the volatile materials like water. Ater that the protoplanetry disk is made of highly volatile materials that exists in solid from only at low coldest temperatures. So the outer part of the protoplanetry disk  is made up of the Methane, Ammonia and Carbon Dioxide.

7 0
3 years ago
Two or more velocities add by ____?<br><br> Plz help
VladimirAG [237]
By vector addition.
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3 0
3 years ago
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65 POINTS! PLEASE ANSWER EVERY QUESTION! NEED HELP ASAP!
otez555 [7]
Maybe you can split up the questions. I will try to answer your first question.

1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)

2. Momentum: p = m₁v₁ + m₂v₂

m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B

Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0

Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0

3: mv + m(-v) = m(-v) + m(v)
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p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
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5.You figure out.



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Hello


The correct answer is A. Jupiter


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Answer:

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Explanation:

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