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KonstantinChe [14]
3 years ago
13

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 diamet

er 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?
Physics
1 answer:
sveta [45]3 years ago
7 0
<span>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</span>
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3 years ago
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4 years ago
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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Alborosie

Answer:

I = 4.75 A

Explanation:

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J=\frac{E}{\rho}      (1)

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The current density is also given by:

J=\frac{I}{A}        (2)

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A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}

Next, you replace for all variables:

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hence, the current in the wire is 4.75A

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