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andrey2020 [161]
3 years ago
8

The tires of a car make 60 revolutions as the car reduces its speed uniformly from 95.0 km/h to 60.0 km/h. The tires have a diam

eter of 0.88 m. If the car continues to decelerate at this rate, how far does it go
Physics
1 answer:
yawa3891 [41]3 years ago
4 0

Answer:

-2.869 rad/s2

Explanation:

Data given:

speed, vi at 95.0 km/h = 95 X (1 hour /3600 seconds) X (1000m / 1km)

Note that, for every 1 hour, there will be 60sec X 60sec = 3600 seconds

And for every 1km, there will be 1000m.

So, speed of 95.0 km/h = 26.389 m/s

speed, vi  =  r ω (radius X angular velocity)

 angular velocity, ωi = v/r

ωi = 26.389  m/s ÷ half of 0.88 m diameter

= 59.975 rad/s

decelerating to speed, vf at 60.0 km/h = 60 X X (1 hour /3600 seconds) X (1000m / 1km)

= 16.667m/s

The angular velocity for this speed = 16.667m/s ÷ half of 0.88 m diameter

 = 37.879rad/s

 How far the car goes is equivalent to the angular acceleration which equals to (ωf^2 - ωi^2) ÷ 2θ

= (37.879rad/s)^2 - (59.975 rad/s)^2 ÷ 2 (60 rev X 2π rad/rev)

= -2.869 rad/s2

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As stated in the statement, we will apply energy conservation to solve this problem.

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\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2

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7 0
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Rudiy27

Answer:

The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

Explanation:

To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.

By definition it is known that the conservation of the moment is given by:

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Our values are given by,

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v_{fy}=\frac{5.25}{2}

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Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

6 0
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