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3 years ago

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A 45 kg wagon is being pulled with a rope that makes an angle of 38 degrees with the horizontal. The applied force is 410 N and

the coefficient of kinetic friction between the wagon and the ground is 0.18.

1 answer:

mel-nik [20]3 years ago

8 0

Answer:

7.35m/s²

Explanation:

From the question we are not told what to find but we can as well find the acceleration of the wagon.

According to newton second law of motion

$\sum F_x = ma_x\\Fm - Ff = ma_x\\Fm - \mu R = ma_x\\Fm- \mu mg = ma_x\\$

Fm is the moving force = 410N

$\mu$ is the coefficient of friction = 0.18

m is the mass = 45kg

g is the acceleration dur to gravity = 9.8m/s²

a is the acceleration of the wagon

Substitute the given data into the equation ang get ax

$Fm- \mu mg = ma_x\\410 - (0.18)(9.8)(45) = 45a_x\\410 - 79.38 = 45a_x\\330.62 = 45a_x\\a_x = 330.62/45\\a_x = 7.35m/s^2\\$

Hence the acceleration of the wagon is 7.35m/s²

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4 years ago

What force is required to move 7 M if the work done is 9 J

Murrr4er [49]

Answer:

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2 years ago

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Answer:

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3 years ago

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marishachu [46]

Answer:

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Explanation:

According to the law of conservation of momentum is that the momentum before the collision is equal to the momentum after the collision. In an equation form it would look like this:

M₁V₁+M₂V₂ = M₁V₁'+M₂V₂'

Where:

M₁ = mass of object 1 (kg)

V₁ = velocity of object 1 before the collision (m/s)

V₁' = Final velocity of object 1 after the collision (m/s)

M₂ = mass of object 2 (kg)

V₂ = velocity of object 2 before the collision (m/s)

V₂' = Final velocity of object 2 after the collision (m/s)

According to your problem you have the following given:

M₁ = 5 g = 0.005kg

V₁ = 3 m/s

V₁' = -5m/s (It bounced off so it is going the other direction)

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V₂ = -1 m/s (It is coming from the opposite direction of the 3-ball)

V₂' = ?

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