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3 years ago

9

A 48.9 kg meteor is moving in outer space. If a 8.6 N force is applied opposite the direction of motion, what is the deceleratio

n (in outer space we assume no friction)

1 answer:

nalin [4]3 years ago

3 0

Answer:

The deceleration is 0.18 m/s²

Explanation:

Hi there!

Using Newton´s second law, we can calculate the deceleration:

∑F = m · a

Where:

∑F = the sum of all forces in a given direction.

m = mass of the object.

a = acceleration.

Solving for a:

∑F/m = a

The only force acting on the meteor is the applied force of 8.6 N. So, the acceleration will be:

8.6 N / 48.9 kg = a

a = 0.18 m/s²

The deceleration is 0.18 m/s² or, in other words, the acceleration is -0.18 m/s²

Have a nice day!

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