A2.2 kg ball strikes a wall with a velocity of 7.4 m/s to the left. the ball bounces off with a velocity of 6.2 m/s to the right. if the ball is in contact with the wall for 0.25 s, what is the constant force exerted on the ball by the wall?

The constant force exerted on the ball by the wall is 119.68 N.

Step-by-step explanation:

Consider the provided information.

It is given that the mass of the ball is m = 2.2 kg

The initial velocity of the ball towards left is 7.4 m/s

So the momentum of the ball when it strikes is =

The final velocity of the ball is -6.2 m/s

So the momentum of the ball when it strikes back is =

Thus change in moment is:

The duration of force exerted on the ball t = 0.25 s

Therefore, the constant force exerted on the ball by the wall is:

Hence, the constant force exerted on the ball by the wall is 119.68 N.

the mean and mad

step-by-step explanation:

15%

step-by-step explanation:

use the table and count the number of guesses that has two correct answers.

the following are two correct answers

10010, 01100, 11000

a total of 3 numbers having 2 correct guesses out of 20 questions.

probability of getting 2 correct guesses = the number of questions getting 2 correct answers / total number of questions.

= 3/20

to convert it into percent multiply by 100, we get

= 3/20 * 100

= 15%

the answer is 15%

you.