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3 years ago

1. °C= (°F-32) X 5/9. what will be the answer?

Physics

1 answer:

aleksley [76]3 years ago

3 0

Answer:

F=9/5(°C)+32

Explanation:

Lets take an example

Take 25°C

$\\ \sf\longmapsto \dfrac{9}{5}(25)+32$

$\\ \sf\longmapsto 9(5)+32$

$\\ \sf\longmapsto 45+32$

$\\ \sf\longmapsto 77°F$

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By Newton's second law,

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where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

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A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take f

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(ignore air resistance) (include units and correct number of significant figures)

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