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Triss [41]
3 years ago
10

A flagpole casts a shadow that is 50 feet long. At the same time, a woman standing nearby who is five feet six inches tall casts

a shadow 4 feet long. How tall is the flagpole?
will give extra points if work is shown

Mathematics
2 answers:
Pavel [41]3 years ago
7 0
The flagpole is 70 feet tall

\frac{5.6}{4} = \frac{x}{50}

Divide, 5.6 by 4

1.4 = \frac{x}{50}

Simplify with cross multiplication 1.4 × 50,

70 = x

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey
V125BC [204]3 years ago
3 0
Check the picture below.

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To solve this problem, we have to find the net displacement and the net force and the multiply the dot product together and get the work done.

The work done on moving the object is 27ft*lbs

<h3>Work done in moving the object from point A to point B</h3>

To find the work done on this object, let's find the net force on the object.

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The unit vector of the force is

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F = 4(\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} })\\F= (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } )

The displacement on the object is 7ft through (0,4) to (5, -1)

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\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }

The net displacement will be

7(\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }) = \frac{35}{\sqrt{50} }, \frac{-35}{50}

The work done will be F.d

w = f. d \\

w = (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } ) * \frac{35}{\sqrt{50} }, \frac{-35}{50}\\w = 17.71+ 8.854\\w = 26.567 = 27ft*lbs

The work done on moving the object is 27ft*lbs

Learn more on work done on an object here;

brainly.com/question/26152883

#SPJ1

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