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kati45 [8]
3 years ago
9

a 0.40kg soccer ball approaches a player horizontally with a velocity of 18m/s north. the player strikes the ball and causes it

to move in the opposite direction with a velocity of 22m/s. what impulse was delivered to the ball by the player?
Physics
1 answer:
denis23 [38]3 years ago
4 0
Impulse = Ft = (m)(delta v)
delta v = change in velocity = velocity final - velocity initial.
= -22m/s - +18m/s = -40m/s.
mdeltav = (0.40kg)(-40m/s) = -16kgm/s or -16Ns.

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A car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The diameter of a tire is 80.2 cm. Find the numb
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By using first and third equation of motion, the number of revolutions the tire makes during this motion is 43 rev.

ANGULAR MOTION

Since the car accelerate from rest, initial velocity will be zero.

Given that a car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The following are the given parameters

  • Initial velocity U = 0
  • Final velocity V = 24.3 m/s
  • Time t = 9.1 s

If the diameter of a tire is 80.2 cm, to find the number of revolutions the tire makes during this motion, we must first calculate the distance travelled by the car by using first and third equation of motion of the car.

First equation

V = U + at

Substitute all necessary parameters into the equation.

24.3 = 0 + 9.1a

a = 24.3/9.1

a = 2.67 m/s^{2}

Third Equation of motion

V^{2} = U^{2} + 2aS

Substitute all the necessary parameters

24.3^{2} = 0 + 2 x 2.67 x S

590.49 = 5.34S

S = 590.49 / 5.34

S = 110.58 m.

Given that the diameter of a tire is 80.2 cm,

the radius (r) will be 80.2/2 = 40.1 cm

convert it to meter

r = 40.1/100 = 0.401 m

The Circumference of the tire = 2\pir

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Circumference = 2.52 m

Assuming no slipping, number of revolutions = 110.58/2.52

Number of revolutions = 43.89 rev.

Number of revolutions = 43 rev.

Therefore, the number of revolutions the tire makes during this motion is 43 rev.

Learn more about circular motion here: brainly.com/question/6860269

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