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AleksAgata [21]
3 years ago
5

Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up

from left to right and a force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 2.0 m/s2 . What is the magnitude of the force exerted on the middle cube by the left cube in this case?

Physics
1 answer:
Westkost [7]3 years ago
4 0

Answer:

24 N

Explanation:

m = mass of the cube = 6.0 kg

Consider the three cubes together as one.

M = mass of the three cubes together = 3 m = 3 (6.0) = 18 kg

a = acceleration of the combination = 2 ms⁻²

F = Force applied on the combination

Using Newton's second law

F = ma = (18) (2) = 36 N

F_{L} = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N

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The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

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W = Eq₂

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The magnitude of the second charge is determined by applying Coulomb's law.

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Learn more about electric potential here: brainly.com/question/14306881

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potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below

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<em>Substituting these values into Equation 1</em>

<em>E = 1/2(10)(2)²</em>

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<em>E = 20 Joules.</em>

<em>Therefore the elastic potential energy stored in the bungee cord = 20 J</em>

<em></em>

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{\underline{\green{\textsf{\textbf{ Answer : }}}}}

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