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Studentka2010 [4]
3 years ago
5

Charges q1 and q2 exerts repulsive forces of 10N on each other what is the repulsive force when their separation is decreasing s

o that their final separation is 80% of their initial separation ?
Physics
1 answer:
Brut [27]3 years ago
6 0

Answer:

15.6 N

Explanation:

Coulomb's law:

F = k q₁ q₂ / r²

At distance r:

10 N = k q₁ q₂ / r²

At distance 0.8r:

F = k q₁ q₂ / (0.8r)²

F = k q₁ q₂ / (0.64r²)

F = 10 N / 0.64

F ≈ 15.6 N

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Answer:

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3 years ago
Need help solving this question.
MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

-15+T-W=0\\T-W=15

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

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2 years ago
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Well, the rings surrounding a planet are made out of rock. A ring surrounding the sun would be impossible since the sun can reach more than 27 million degrees Fahrenheit (15 million degrees Celsius.)

Hope this helped.

7 0
3 years ago
4.   Which of the following is the term used to describe a body's resistance to a change in motion? 
Alborosie
Inertia ..................
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