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krek1111 [17]
2 years ago
9

ox A has a mass of 20.0kg and Box B has a mass of 30.0kg. Box A is placed on a horizontal surface that is frictionless and Box B

is hanging from a pulley by a rope connected to Box A. If the acceleration of the system is 5.88m/s2, what is the Tension T on Box A?
Physics
1 answer:
MatroZZZ [7]2 years ago
3 0

The tension on box A is 117.6 N

<h3 /><h3>What is Equilibrium of Forces ?</h3>

Equilibrium is a state of an object with respect to a given observable quantity during the time for which there is no change in that quantity.

A body is said to be in equilibrium when:

  • the body as a whole either remains at rest or moves in a straight line with constant speed.
  • the body is either not rotating at all or is rotating at a constant angular velocity.

Given that box A has a mass of 20.0kg and Box B has a mass of 30.0kg. Box A is placed on a horizontal surface that is frictionless and Box B is hanging from a pulley by a rope connected to Box A.

The tension in the rope will be the same.

Given that Box A is placed on a horizontal surface that is frictionless, that is,

T = ma .... (1)

Also,  Box B is hanging from a pulley by a rope connected to Box A. That is,

W - T = ma ...... (2)

Where W = mg

Since the acceleration of the system is 5.88m/s² which will be the same for the two equations, then the Tension T on Box A can be calculated by substituting acceleration and mass into equation (1)

T = 20 x 5.88

T = 117.6 N

Therefore, the tension on box A is 117.6 N

Learn more about Equilibrium of Forces here: brainly.com/question/517289

#SPJ1

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Answer:

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Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges q_1 and q_2 separated by a distance d, the magnitude of the electrostatic force F between them is:

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Taking the values:

d = 4.6 \ 10^{-15} m

q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}

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Taking all this in consideration:

F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}

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Answer:

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