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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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3 years ago
A jet airliner moving initially at 889 mph
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Answer:

1500 mph

Explanation:

Take east to be +x and north to be +y.

The x component of the velocity is:

vₓ = 889 cos 0° + 830 cos 59°

vₓ = 1316.5 mph

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vᵧ = 889 sin 0° + 830 sin 59°

vᵧ = 711.4 mph

The speed is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1316.5 mph)² + (711.4 mph)²

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Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.

8 0
3 years ago
You shine a LASER from the air into water. What happens to light?
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The speed of an electric locomotive is 90kmh .express this speed in m/s with method​
Maksim231197 [3]

Answer:

24.3 m/s

Explanation:

1 kmh = 0.27 m/s, that makes a conversion ratio of 0.27/1kmh

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The "kmh" n the top and bottom cancel out. And  then you just multiply the top 90 x 0.27 and the bottom 1 x 1 to get

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A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca
igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

f=mg\ sin\theta

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f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0
3 years ago
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