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Firdavs [7]
2 years ago
10

A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with

Physics
2 answers:
wariber [46]2 years ago
5 0

From Newton's second law of motion, the average force on the ball by the racket is 98 Newtons. The correct answer is option C

Given that a 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. And the time for contact with the racket is 0.04 s, that is,

mass m = 0.07 kg

velocity v = 56 m/s

time t = 0.04 s

force f = ?

To calculate the average force on the ball by the racket, let us apply Newton's second law of motion.

Impulse = change in momentum

ft = mv

Substitute all the parameters into the equation above

0.04f = 0.07 x 56

make f the subject of the formula

f = 3.92 / 0.04

f = 98 N

Therefore, the average force on the ball by the racket is 98 Newtons. The correct answer is option C

Learn more about momentum here: brainly.com/question/7538238

Naddika [18.5K]2 years ago
4 0

The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

From the question, we are to determine the average force on the ball by the racket.

From the formula,

F = \frac{mv}{t}

Where F is the force

m is the mass

v is the velocity

and t is the time

From the given information

m = 0.07 kg

v = 56 m/s

t = 0.04 s

Putting the parameters into the formula,

we get

F = \frac{0.07 \times 56}{0.04}

F = \frac{3.92}{0.04}

F = 98 N

Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

Learn more on calculating force exerted on an object here: brainly.com/question/13590154

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Answer:

Phase difference = pi/4 radians

Explanation:

Given:

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Find:

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Solution:

- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.

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- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.

                                  r = sqrt ( L^2 + 0.0025^ )

Where, r is the distance from split to the interference bright fringe.

                                  r = sqrt(2.8^ + 0.0025^) = 2.8

                                  sin(Q) = 0.0025 / 2.8

Hence.                        m = 0.00085*0.0025 / 2.8*(600*10^-9)

                                   m = 1.26

- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.

- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).

Answer:                  Phase difference = pi/4 radians

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