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Crank
1 year ago
15

a woman walks south at a speed of 2.0mph for 60 minutes. She then turns around and walks north at a distance of 3000m in 25 minu

tes. What is the woman's average speed during her entire motion?
Physics
1 answer:
TiliK225 [7]1 year ago
8 0

When a  woman walks south at a speed of 2.0mph for 60 minutes. She then turns around and walks north at a distance of 3000m in 25 minutes. then the woman's average speed during her entire motion would be  73.15 meters /minute.

<h3>What is speed?</h3>

The total distance covered by any object per unit of time is known as speed.

the mathematical expression for speed is given by

speed = total; distance /total time

As given in the problem a woman walks south at a speed of 2.0mph for 60 minutes

60 min = 1 hour

1 mile = 1.60934 km

The distance covered by her southwards  =  speed ×time

                                                                      =2 mph × 60 minutes

                                                                      = 3.218 km

She then turns around and walks north at a distance of 3000m in 25 minutes

The distance covered northward is 3000m

speed = total distance /total time

          =(3218 +3000) /(60+25)

          =73.15 meters /minutes

Thus, The average speed of the woman would be 73.15 meters /minute.

Learn more about speed from here

brainly.com/question/13263542

#SPJ1

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Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

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Let's use the parabolic motion equation to solve it. Let's define the variables:

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For the first stone

y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}              

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If we solve the equation (1) we will have:

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We can do the same procedure for the equation (2)

t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

x_{f1}=x_{i1}+vcos(30)t_{1}

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<u>Second stone</u>

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x_{1}=0+13.86*t_{1}

x_{1}=13.86*t_{2}

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

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I hope it helps you!

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