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hichkok12 [17]
3 years ago
14

A distressed car is rolling backward, downhill at 3.0 m/s when its driver finally manages to

Physics
1 answer:
konstantin123 [22]3 years ago
5 0

Answer:

Explanation:

Acceleration is equal to the change in velocity over the change in time, or

a=\frac{v_f-v_i}{t} where the change in velocity is final velocity minus initial velocity. Filling in:

3.0=\frac{v_f-(-3.0)}{6.0} Note that I made the backward velocity negative so the forward velocity in our answer will be positive.

Simplifying that gives us:

3.0=\frac{v_f+3.0}{6.0} and then isolating the final velocity, our unknown:

3.0(6.0) = v + 3.0 and

3.0(6.0) - 3.0 = v and

18 - 3.0 = v so

15 m/s = v and because this answer is positive, that means that the car is no longer rolling backwards (which was negative) but is now moving forward.

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To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

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At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

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Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

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\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

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\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

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3 years ago
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Answer:

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White raven [17]

Answer:

here given is a weight

then force becomes mg

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