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ELEN [110]
2 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]2 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]2 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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SIZIF [17.4K]

Answer:

\frac{A_{slow} }{A_{fast} }=6.2885

Explanation:

Given data

Terminal velocity for spread eagle position vt=130 km/h

Terminal velocity for nosedive position vt=326 km/h

The terminal speed of the diver is given by

v_{t}=\sqrt{\frac{2mg}{CpA} }

Therefore the area is given by

A=\frac{2mg}{Cp(v_{t})^{2}  }\\

Since everything else is constant in the two dives except for the terminal velocity, the ratio between the area in the slow position to the area in the fast position is

\frac{A_{slow} }{A_{fast} }=\frac{326^{2} }{130^{2} }\\\frac{A_{slow} }{A_{fast} }=6.2885

8 0
3 years ago
What is the ratio of the sun’s gravitational pull on Mercury to the sun’s gravitational pull on the earth?
Marta_Voda [28]

Answer:

The answer is \frac{F_{Sun-Mercury} }{F_{Sun-Earth} } =0,3709. Let's learn why.

Explanation:

Newton's law of universal gravitation says;

F_{g} =G.\frac{m_{1}.m_{2}}{r^{2}}

Here G is a universal gravitational <u>constant</u> and is measured experimentally.

Sun's gravitational pull on mercury is:

F_{Sun-Mercury} =G.\frac{m_{sun}.3,30.10^{23}}{(5,79.10^{10})^{2} }

Therefore F_{Sun-Mercury} = Gm_{sun} 98,4366

Sun's gravitational pull on Earth is:

F_{Sun-Earth} =G.\frac{m_{sun} 5,97.10^{24} }{(1,50.10^{11}) ^{2}}

Therefore F_{Sun-Earth} =Gm_{sun} 265,33

As a result;

\frac{F_{Sun-Mercury}}{F_{Sun-Earth} }=\frac{Gm_{sun}98,4366}{Gm_{sun}265,33 } =0,3709

4 0
3 years ago
The resolution of a camera or other optical system is determined by the relationship between what two scales?
devlian [24]

Answer:

d.The wavelength of light and the size of the aperture

Explanation:

<em>The resolution power of an optical system is the smallest distance between two points that the device can distinguish clearly.</em>

It has the following relationship:

r=\frac{\lambda}{2n}

where:

r = minimum resolvable distance

n = numerical aperture

\lambda= wavelength of the light used for viewing

From above mathematical equation it is clear that:

  • Smaller the wavelength better the resolving power
  • Larger the aperture better the resolution

(Note, that smaller the value of "r" the more finer details of the image visible through the device.)

4 0
2 years ago
Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five second
REY [17]

Answer:

E = 75 J

Explanation:

First, we will calculate the total power consumed by the five lamps:

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Now, the energy supply can be calculated as follows:

E = Pt

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E = Energy = ?

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Dafna1 [17]
The answer is the dew point
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