The tires of a car make 75 revolutions as the car reduces its speed uniformly from 95 km/hr to 55 km. hr. the tires have

The tires of a car make 75 revolutions as the car reduces its speed uniformly from 95 km/hr to 55 km. hr. the tires have a diameter of 0.80 m. (a) what was the angular acceleration of the tires? (b) if the car continues to decelerate at this rate, how much more time is required for it to stop and (c) how far does it go?

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  1. 95 km/h = 26.39 m/s (95000m/3600 secs) 55 km/h = 15.28 m/s (55000m/3600 secs) 75 revolutions = 75 x 2pi = 471.23 radians radius = 0.80/2 = 0.40m v/r = omega (rad/s) 26.39/0.40 = 65.97 rad/s 15.28/0.40 = 38.20 rad/s s/((vi + vf)/2) = t 471.23 /((65.97 + 38.20)/2) = 9.04 secs (vf - vi)/t = a (38.20 - 65.97)/9.04 = -3.0719 The angular acceleration of the tires = -3.0719 rad/s^2 Time is required for it to stop (0 - 38.20)/ -3.0719 = 12.43 secs How far does it go? 65.97 - 38.20 = 27.77 M

  2. answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

    explanation:

    the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

    thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

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