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Ede4ka [16]
4 years ago
8

A bug lands on a windshielf wiper. Explain why the bug is more likely to be dislodged when the wipers are turned on at the high

rather than low setting.
Physics
1 answer:
GREYUIT [131]4 years ago
6 0

Answer and Explanation:

The motion of the viper is circular and it completes a semi- circle forming an arc and then retraces its path back.

Thus the force experienced by the bug is centripetal force thus centripetal acceleration and in the absence of this force the bug will get dislodged.

This centripetal force is mainly provided by the static friction between the blades and the bug.

When the wipers are moving with high velocity when turned on, larger centripetal force is required to keep the bug moving with the wiper on the arc  than at low setting.

Thus there are more chances for the bug to be dislodged at higher setting than at low setting.

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A motorboat accelerates uniformly from a velocity of 6.5 m/s west to a velocity of 1.5 m/s west. If its acceleration was 2.7 m/s
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Answer:

<em>The motorboat ends up 7.41 meters to the west of the initial position </em>

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<u>Accelerated Motion </u>

The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:

\vec v_f=\vec v_o+\vec a.t

\displaystyle \vec r=\vec v_o.t+\frac{\vec a.t^2}{2}

The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is 2.7 m/s^2 to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas:

v_f=v_o+a.t

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Solving the first one for t

\displaystyle t=\frac{v_f-v_o}{a}

We have

v_o=-6.5,\ v_f=-1.5,\ a=2.7

Using these values

\displaystyle t=\frac{-1.5+6.5}{2.7}=1.852\ s

We now compute x

\displaystyle x=(-6.5)(1.852)+\frac{(2.7)(1.852)^2}{2}

x=-7,41\ m

The motorboat ends up 7.41 meters to the west of the initial position

5 0
3 years ago
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