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2 years ago

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A crane lifts a 425 kg steel beam vertically a distance of 66 m. How much work does the crane do on the beam if the beam acceler

ates upward at 1.8 m/s2? Neglect frictional forces ...?

2 answers:

Fiesta28 [93]2 years ago

6 0

<span>The solution to the problem is as follows:

W=Fd=mad=425*1.8*66 =50490 J

Therefore, it takes </span><span>50490 J of work the crane will do to accelerate the beam upward at 1.8 m/s^2.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

</span>

Guest1 year ago

0 0

425 x 66 x 9.81 = 275170.5 J = 2.75 X 10 ^5 J = 275.17 kJ

Guest

1 year ago

then, minus 425 x 1.8 x 66 = 224 680.5 J

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2 years ago

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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

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Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

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$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

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Answer:

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However, since 1.5 times as many men voted as women, we have to apply our poll to 15 men and 5 women.

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67% of the 5 women makes 3.33 women who voted for Smith and 1.67 who did not.

Add these numbers together.

8.33 total voters were cast for John Smith, while 11.67 were not.

Now, divide 8.33 by 20 and multiply by 100 to obtain a percentage.

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2 years ago