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3 years ago

If the force acting on a cart doubles, what happens to the carts acceleration?

Physics

1 answer:

Effectus [21]3 years ago

8 0

Answer:

If the force on a cart doubles, the acceleration of the cart doubles.

Explanation:

For this problem, we need to consider the following equation:

Force = Mass x Acceleration

We can reasonably assume that the cart will have constant mass in the given force system. With this assumption we can say the following relationship:

Force is directly proportional to Acceleration within the system.

Given that our force on the cart is doubled, then our acceleration of the cart must also be doubled. You can mathematically express this as follows:

F = MA

2F = M * 2A

Hence, if force doubles, the acceleration doubles.

Cheers.

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3 years ago

Read 2 more answers

A 3.0 kg block is pushed by a 14 N force. If µ = 0.6, will the block move?

Anna71 [15]

Answer:

The block will not move.

Explanation:

We'll begin by calculating the frictional force. This can be obtained as follow:

Coefficient of friction (µ) = 0.6

Mass of block (m) = 3 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (R) = mg = 3 × 10 = 30 N

Frictional force (Fբ) =?

Fբ = µR

Fբ = 0.6 × 30

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From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.

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2 years ago

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

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$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

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3 years ago

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Answer:

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2 years ago

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Answer:

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