How much force is needed to stop a 90-kg soccer player if he decelerates at 15 m/s^2?

1 answer:

8 0

We have: F = m×a
Here, m = 90 Kg
a = 15 m/s²

Substitute their values into the expression:
F = 90 × 15
F = 1350 N

In short, Your Answer would be Option D

Hope this helps!

You might be interested in

A certain automobile is 6.0 m long if at rest. If it is measured to be 4.8 m long while moving, its speed is:

tatuchka [14]

Answer:

1.8*10^8m/s

Explanation:

Using

L= lo√1-v²/c²

So making v subject we have

V= c√1-4.8²/6²

V= 0.6*c

V= 0.6*3E8m/s

V= 1.8*10^8m/s

7 0

2 years ago

The mass of a proton is approximately equal to

Dahasolnce [82]

Answer:

$1.6726\cdot 10^{-27} kg$

Explanation:

The three main particles that make an atom are:

- Proton: its mass is $1.6726\cdot 10^{-27} kg$ , it carries an electric charge of +e ( $e=1.6\cdot 10^{-19}C$ ), and it is located in the nucles of the atom

- Neutron: its mass is $1.6749 \cdot 10^{-27}kg$ , it carries no electric charge, and it is also located in the nucleus of the atom

- Electron: its mass is $9.1094 \cdot 10^{-31}kg$ , it carries an electric charge of -e ( $e=1.6\cdot 10^{-19}C$ ), and it is located outside the nucleus

3 0

3 years ago

An 80kg astronaut traveled to the moon, where gravity is one-sixth (116) as

PIT_PIT [208]

Answer:

Wmoon = 131 [N]

Explanation:

We know that the weight of a body is equal to the product of mass by gravitational acceleration.

Since we are told that the gravitational acceleration of the moon is equal to one-sixth of the acceleration of Earth's gravitation. Then we must multiply the value of Earth's gravitation by one-sixth.

$w_{moon}=\frac{1}{6} *m*g\\w_{moon}=\frac{1}{6} *80*9.81\\w_{moon}=130.8 [N] = 131 [N]$