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Sergeeva-Olga [200]
3 years ago
10

An artificial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Dete

rmine the orbital period of the satellite.
Physics
1 answer:
Ksju [112]3 years ago
6 0
 <span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:


R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km 

here, the speed of the satellite is:

v = sqrt(R*9.00m/s^2) = 7730 m/s 

  the time it would take the satellite to complete one full rotation is:

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h 

Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu!  :)</span>
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A 6.2 kg ladder, 1.97 m long, rests on two sawhorses. Sawhorse A is 0.64 m from one end of the ladder, and sawhorse B is 0.17 m
Keith_Richards [23]

Answer:

42.69 N and 18.07 N

Explanation:

We are given that

Mass of ladder=6.2 kg

Length of ladder=1.97 m

Distance of Sawhorse A from one end=0.64 m

Distance of sawhorse B from other end=0.17 m

Let center of Ladder=\frac{1.97}{2}=0.985 m

Now, the distance of sawhorse A from center=r=0.985-0.64=0.345 m

Distance of sawhorse B from center of ladder=0.985-0.17=0.815  m

Force one ladder due to gravity=mg=6.2\times 9.8=60.76N

Where g=9.8 m/s^2

Torque applied on Sawhorse A=0.345F_a

Torque applied on Sawhorse B=0.815F_b

In equilibrium

0.345F_a=0.815F_b

F_b=\frac{0.345}{0.815}F_a

Total force=F_a+F_b

F_a+\frac{0.345}{0.815}F_a=60.76

\frac{0.815F_a+0.345F_a}{0.815}=60.76

\frac{1.16}{0.815}F_a=60.76

F_a=\frac{60.76\times 0.815}{1.16}=42.69 N

F_b=\frac{0.345}{0.815}\times 42.69=18.07 N

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