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dusya [7]
2 years ago
8

If you are following a car and also being tailgated by another vehicle, your best option would be to.. a. Increase the following

distance between you and the car in front of you b. Slam on your breaks immediately c. Speed up to increase the space between you and the tailgating car d. Slowly and steadily apply your breaks
Physics
2 answers:
Pie2 years ago
7 0

Answer:

Your answer here is D

Explanation:

Slowly pressing your breaks will help ensure you are not hit by the other car. If they hit you its their fault. Hope this helps :)!

Rufina [12.5K]2 years ago
5 0

Answer:

<h2>D. Slowly and steadily apply your breaks.</h2>

Explanation:

When a person is being tailgated by another vehicle there's a risk of accident, due to the difference of speeds. When someone is being tailgated by another vehicle, it's because they vehicle doesn't work, cannot move by itself, so one vehicle impulse another.

To avoid any accident, the vehicle that's being tailgated must apply breaks often and slowly to try to maintain the same constant distance and speed than the other vehicle. This constant distance is what avoid accidents.

Therefore, the right answer is D.

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20N•m or 20J. Work is equal to force•distance, and 5N•4m is 20N•m, or J
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The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo
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Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s

The tire angular velocity before stopping is:

\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s

Also its angular decceleration:

\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2

Using the following equation motion we can findout the angle it makes during the deceleration:

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where \omega = 0 m/s is the final angular velocity of the car when it stops, \omega_0 = 114rad/s is the initial angular velocity of the car \alpha = 14.75 rad/s2 is the deceleration of the can, and \Delta \theta is the angular distance traveled, which we care looking for:

-114^2 = 2*(-14.75)*\Delta \theta

\Delta \theta = 440rad or 440/2π = 70 revelutions

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For the clay, it final velocity is zero since it sticks to the floor, hence (v =0)

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