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KengaRu [80]
2 years ago
14

Lora (of mass 54 kg) is an expert skier. She

Physics
1 answer:
omeli [17]2 years ago
6 0

The mechanical energy at top =Mechanical energy at bottom

  • Mass=m=54kg
  • Height=h=51m
  • Acceleration due to gravity=g=10m/s^2
  • Velocity=v=2.6m/s

\\ \tt\longmapsto M_{initial}=M_{Final}

  • Final energy at bottom=The kinetic energy

\\ \tt\longmapsto KE=M_{initial}

\\ \tt\longmapsto KE=P.E_{(Top)}+K.E_{(Top)}

\\ \tt\longmapsto K.E=mgh+\dfrac{1}{2}mv^2

\\ \tt\longmapsto K.E=m\left(gh+\dfrac{v^2}{2}\right)

\\ \tt\longmapsto K.E=54\left((10)(51)+\dfrac{2.6^2}{2}\right)

\\ \tt\longmapsto K.E=54\left(510+\dfrac{6.76}{2}\right)

\\ \tt\longmapsto K .E=54(510+3.38)

\\ \tt\longmapsto K.E=54(513.38)

\\ \tt\longmapsto K.E=27722.52J

\\ \tt\longmapsto K.E=27.7KJ

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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
BaLLatris [955]

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

Circumference = 2 x 3.143 x 0.401

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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2 years ago
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