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mel-nik [20]
3 years ago
6

During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s. What was the ball's speed 2.2 s after she kicked

it?
Physics
1 answer:
ki77a [65]3 years ago
5 0

<u>Answer</u>

3.44 m/s


<u>Explanation</u>

The motion apply the equations of Newton's law of motion. The ball is acceleration is -9.8 m/s² (acceleration due to gravity. It is negative because the ball is going against gravity, so it is decelerating).

The first equation of Newton's law of motion is;

V = U + at

Where V is the final velocity, U is the initial velocity, a is acceleration and t the time taken.

V = 25 + (-9.8 × 2.2)

    = 25 - 21.56

     = 3.44 m/s

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An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be
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Explanation:

The given data is as follows.

    Inner wheel Radius = 20 m,

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where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = \frac{u}{(R - d)}

                  u = V \times \frac{(R - d)}{R}

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If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

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Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

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So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

                   = \frac{V}{r} \times 0.1651

5 0
4 years ago
Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet
Svetllana [295]
Given:

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The part of the light that bounces away from the boundary and heads back
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