You can choose five (5) out of seven (7) questions. You may work the problems in

Question

You can choose five (5) out of seven (7) questions. You may work the problems in any order. Start each problem on a new page. Number your solutions as the problems are numbered here. A correct answer will be considered wrong unless you SHOW ALL WORK. A 400 N sphere of 0.20 m in radius rolls without slipping 5.0 m down a ramp that is inclined at an angle of 30 degree with the horizontal. Use the conservation of energy to find the speed of the sphere at the bottom of the slope if it starts from rest (I = 2/5 MR^2) A) A satellite moves in a circular orbit around earth at a speed of 2, 500 m/s determine a. The satellite altitude above the surface of Earth. b. The period of the satellites orbit. (B) What is the orbital speed of a satellite in a geosynchronous circular orbit 3.58 times 10^7 meters above the surface of the Earth? A turntable with a moment of inertial of 4 times 10^-3 kg middot m^2 rotates freely with an angular speed of 20 revolutions per second. Riding on the rim of the turntable, 15 cm from the center is a cute, 50 g grams mouse. Calculate the angular speed of the turntable when the mouse reaches the distance of center. Two mass, m_1 = 15 kg, m_2 = 3 kg in Atwood's machine are connected over the frictionless pulley that is a uniform disk with a radius of 12 cm. Mass of the pulley is 0.2 kg. Use conservation of energy to solve for the final speed of block m_1 and m_2, when block m_2 moves down a distance of 1.2 m. uniform rod of negligible mass has a sign of mass of 50 kg is hung off its end as shown the diagram If theta = 30.0 degree, determine the tension in the cable and the vertical and horizontal components reaction force at the pin.

Explanation / Answer

For a circular orbit, here are two basic equations.
v = ( G M / r )
T = 2r / v
where r is the orbit radius of the satellite and T is the period

The Earth's mass = 5.97x1024 kg  

G = 6.67e-11 m³/(kg s²)

R=6.378 x 106m

v = 2500m/s

(a) We get
r = G M / v² = 6.37 x 107m

Since Earth's radius is 6.378 x 106m

Thus, the satellite's altitude above the Earth's surface is

6.37 x 107 - 6.378 x 106m = 5.7 x 107 m

(b)
T = 2r / v = 1.6 x 105 s = 44.4 h

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