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KatRina [158]
3 years ago
7

A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 358

80Km above the earth surface. Find the period of revolution of the satellite. (Take the mass of earth =5.98×10²⁴kg, the radius of the earth 6370km and G=6.6×10–¹¹Nm²/kg2)​
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0

Answer:

Period is 86811.5 seconds.

Explanation:

{ \boxed{ \bf{T {}^{2} =  (\frac{4 {\pi}^{2} }{GM}) {r}^{3}   }}}

{ \tt{T {}^{2}  =  \frac{4 {(3.14)}^{2} }{(6.6 \times  {10}^{ - 11} ) \times (5.98 \times  {10}^{24} )} \times  {((35880\times  {10}^{3}) } + (6370 \times  {10}^{3} )) {}^{3}   }} \\  \\ { \tt{T {}^{2}  =  7.54 \times {10}^{9} }} \\ { \tt{T =  \sqrt{7.54 \times  {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}

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On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.6 cm and diameter 2.95 cm fr
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Answer:

48.4293354946 N

Yes

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d = Diameter of rod = 2.95 cm

h = Length of rod = 92.6 cm

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g = Acceleration due to gravity = 9.81 m/s²

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V=\dfrac{1}{4}\pi d^2h\\\Rightarrow V=\dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}

Mass is given by

m=\rho V\\\Rightarrow m=7800\times \dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}\\\Rightarrow m=4.93673144695\ kg

Weight is given by

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The weight of the rod is 48.4293354946 N

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7 0
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However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.


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A spring stretched for 1.67 m distance.

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