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oksian1 [2.3K]
2 years ago
5

A force does 35000J of work along a distance of 10.5m. Find the applied force.

Mathematics
1 answer:
nadezda [96]2 years ago
3 0

Answer: 3333.333 N

Work: To figure it out, you must do 35000J divided by 10.5. After diving, you receive the answer 3333.333N. I hope this helps you!

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What is the result of a dilation of scale factor 3 centered at the origin of the line 2y + 3x=10?? PLEASE HELP PLEASEEEEEEEEE
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Given:

The equation of a line is:

2y+3x=10

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To find:

The result of dilation.

Solution:

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2y+3x=10

For x=0,

2y+3(0)=10

2y+0=10

y=\dfrac{10}{2}

y=5

For x=2,

2y+3(2)=10

2y+6=10

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y=\dfrac{4}{2}

y=2

The given line passes through the two points A(0,5) and B(2,2).

If the line dilated by factor 3 with origin as center of dilation, then

(x,y)\to (3x,3y)

Using this rule, we get

A(0,5)\to A'(3(0),3(5))

A(0,5)\to A'(0,15)

Similarly,

B(2,2)\to B'(3(2),3(2))

B(2,2)\to B'(6,6)

The dilated line passes through the points A'(0,15) and B'(6,6). So, the equation of dilated line is:

y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)

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y-15=\dfrac{-3}{2}x

Multiply both sides by 2.

2(y-15)=-3x

2y-30=-3x

2y+3x=30

Therefore, the equation of the line after the dilation is 2y+3x=30.

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