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densk [106]
3 years ago
6

You are diving a car out in rural America and suddenly you get a flat tire. you pull off to the side of the road. you check your

cell reception and dont have a single bar of signal (your cell phone won't work). you look around and there is no other cars or houses anywhere in sight and you do not remember the last time you saw a passing car or house. you are absolutely alone. the sun is setting and it is just starting to get dark. you get out and jack up the car, loosen the 4 nuts, take off the wheel with the flat tire. you grab the spare from the trunk and slide it onto the axle- you then look around for the 4 nuts and they are no where to he found. you search everywhere and cant find them. its gettint cold and dark. you look at the car, what can you do? all four of the nuts from the wheel have vanished snd are just gone. explain what you can do and why? remember, you cant call anyone because there is no reception, there is no one else on the road and you havent seen anyone in hours, its getting dark and cold, the 4 nuts from the wheel are gone, the car cannot drive safely with only 3 wheels​
Engineering
1 answer:
julsineya [31]3 years ago
4 0

Answer:

Walk to saftey!

Explanation:

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The distance below the top of the cliff that the two balls cross paths is 7.53 meters.

<u>Given the following data:</u>

  • Initial velocity = 0 m/s (since the ball is dropped from rest).
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<u>Scientific data:</u>

  • Acceleration due to gravity (a) = 9.8 m/s^2.

To determine how far (distance) below the top of the cliff that the two balls cross paths, we would apply the third equation of motion.

<h3>How to calculate the velocity.</h3>

Mathematically, the third equation of motion is given by this formula:

V^2 = U^2 +2aS

<u>Where:</u>

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  • U is the initial velocity.
  • a is the acceleration.
  • S is the distance covered.

Substituting the parameters into the formula, we have;

V^2 = 0^2 +2(9.8) \times 30\\\\V^2 = 588\\\\V=\sqrt{588}

V = 24.25 m/s.

<u>Note:</u> The final velocity of the first ball becomes the initial velocity of the second ball.

The time at which the two balls meet is calculated as:

Time = \frac{S}{U} \\\\Time = \frac{30}{24.25}

Time = 1.24 seconds.

The position of the ball when it is dropped from the cliff is calculated as:

y_1 = h-\frac{1}{2} at^2\\\\y_1 = 30-\frac{1}{2} \times 9.8 \times 1.24^2\\\\y_1 = 30-7.53\\\\y_1=22.47\;meters

Lastly, the distance below the top of the cliff is calculated as:

Distance = 30-22.47

Distance = 7.53 meters.

Read more on distance here: brainly.com/question/10545161

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