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ELEN [110]
3 years ago
6

A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. W

hat are the smallest and largest areas that the field could be?
Physics
2 answers:
tatyana61 [14]3 years ago
5 0

Answer:

The smallest and largest areas could be 6400 m and 7500 m, respectively.

Explanation:

The area of a rectangle is given by:

A = l*w

Where:

l: is the length = 100 m

w: is the width

We can calculate the smallest area with the lower value of the width.

A_{s} = 100 m*64 m = 6400 m^{2}                            

And the largest area is:

A_{l} = 100 m*75 m = 7500 m^{2}  

Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.            

I hope it helps you!                        

Bezzdna [24]3 years ago
4 0

Answer:

the largest areas that the field could be is A_l=7587.75 m

the smallest areas that the field could be is A_s=6318.25 m

Explanation:

to the find the largest and the smallest area of the field measurement error is to be considered.

we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.

therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:

A_l= (L+0.5)(W+0.5)

(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m

To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:

A_s= (L-0.5)(W-0.5)

(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m

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Answer:

Δθ = 15747.37 rad.

Explanation:

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       \theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)

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       \omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)

  • Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

      \theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)

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  • Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad
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