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2 years ago

15

Suppose a force of 100 N is used to push an object a distance of 5 meters in 15 seconds: Find the done and the power for this si

tuation.

1 answer:

zubka84 [21]2 years ago

5 0

Answer:

Step-by-step explanation:

Remark

You need the work done (I think), and the power developed from the givens. This question is just a straight forward application of the formulas for work and power.

Givens

F = 100 N

d = 5 meters

t = 15 seconds

Formulas

W = F * d

Power = F * d / t

Solution

<u><em>Work</em></u>

W = 100 N * 5 meters

W = 500 Newton Meters or 500 J

Work = 500 Joules

<u><em>Power</em></u>

P = F * d / t

Power = 100 * 5 / 15

Power = 33.3 Watts.

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Taking square root on both sides.

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Using trigonometric ratio formula:

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$$\cos \theta=\frac{\text { adjacent side }}{\text { hypotenuse }}$

$$\cos \theta=\frac{10}{18}$

$$\sec \theta=\frac{\text { hypotenuse }}{\text { adjacent side }}$

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$$\tan \theta=\frac{4\sqrt{14} }{10}$

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An arched drainage culvert can be modelled by the function:

Zanzabum

Substitute 136.5 for y.

$136.5=\frac{x(191-x)}{60}$

$8190=x(191-x)$

$8190=191x-x^{2}$

$x^{2} -191x+8190=0$

Compare this equation with $ax^{2} +bx+c=0$

a = 1, b = - 191, c = 8190

$x=\frac{-b+\sqrt{b^{2}-4ac } }{2}$ or

$x=\frac{-b-\sqrt{b^{2}-4ac } }{2}$

$x=\frac{191+\sqrt{191^{2}-4(1)(8190) } }{2}$ or

$x=\frac{191-\sqrt{191^{2}-4(1)(8190) } }{2}$

$x=\frac{191+\sqrt{36481-32760 } }{2}$ or

$x=\frac{191-\sqrt{36481-32760 } }{2}$

$x=\frac{191+\sqrt{3721 } }{2}$ or

$x=\frac{191-\sqrt{3721 } }{2}$

$x=\frac{191+61}{2}$ or

$x=\frac{191-61}{2}$

$x=\frac{252}{2}$ or

$x=\frac{130}{2}$

x = 126 or x = 65

Hence, the points on the curve are (65, 136.5) and (126, 136.5).

The width of the air space is the distance between these points.

Width = $\sqrt{(x_{2}- x_{1})^{2} +( y_{2}- y_{1}) ^{2}$

= $\sqrt{(126- 65)^{2} +(136.5-136.5) ^{2}$

= 126 - 65

= 61

Hence, width of the air space is 61m.

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