1) a star follows a circular orbit around a supermassive black hole in the cente

Question

1) a star follows a circular orbit around a supermassive black hole in the center of a galaxy at a distance of 1 light year with an orbital velocity of 3750 km/s. What is the mass of the black hole in solar mass units?

2) A .5kg mass attached to a spring with a force constant of 8N/m vibrates in simple harmonic motion with an amplitude of 10cm. Calculate a) the maximum value of its speed and acceleration. b) the speed and acceleration of the mass it it is 6.0cm from the equilibrium position and c) the time interval required for the object to move from x=0.0cm to x=8cm

3) A baseball outfielder throws a 0.2kg baseball at a speed of 50m/s and an initial angle of 30 degrees (ignore air resistance and the height of the outfielder). What is a) the kinetic energy of the ball when it hits the ground b)the kinetic energy at the highest point c)the potential energy at its highest point, d) its total time in the air

Explanation / Answer

(3) mass of the ball m = 0.2 kg initial speed of the ball u = 50 m/s initial angle = 300 ............................................. (A) time taken by the ball before reaches the ground T = 2usin/g    = 2*(50 m/s)(sin300) / 9.8 m/sec2    = 5.10 sec velocity at this point vx = ucos      = (50 m/s)(cos 300)      = 43.30 m/s vy = usin-gt       = (50 m/s)(sin 300) - (9.8 m/sec2)(5.10 sec)        = -24.98 m/sec2 resultant velocity V = sqrt(vx2+vy2)                                 = sqrt (2498.89)                                  = 49 m/sec kinetic energy K.E = 1/2 m v2                               = 1/2 (0.2 kg)(49 m/sec)                                = 4.99 J ................................................................................................... (b) At maximum height the vertical velocity is zero and horizontal velocity vx = ucos                                                                                                                       = 43.30 m/s v = sqrt(43.30m/s)    = 6.5 m/s Kinetic energy   = 1/2 m v2                               = 1/2 (0.2 kg)(6.5 m/sec)                                = 0.65 J (c) Potential energy u = mgh    h is the maximum height = u2sin2/2g                                = (0.2 kg)(9.8 m/sec2)[(50m/s)2(sin300)2/2*9.8 m/sec2]                                = 125 J (d) time taken by the ball before reaches the ground T = 2usin/g    = 2*(50 m/s)(sin300) / 9.8 m/sec2    = 5.10 sec NOTE: YOU SHOULD NOT POST MULTIPLE QUESTIONS.                               = 1/2 (0.2 kg)(49 m/sec)                                = 4.99 J ................................................................................................... (b) At maximum height the vertical velocity is zero and horizontal velocity vx = ucos                                                                                                                       = 43.30 m/s v = sqrt(43.30m/s)    = 6.5 m/s Kinetic energy   = 1/2 m v2                               = 1/2 (0.2 kg)(6.5 m/sec)                                = 0.65 J (c) Potential energy u = mgh    h is the maximum height = u2sin2/2g                                = (0.2 kg)(9.8 m/sec2)[(50m/s)2(sin300)2/2*9.8 m/sec2]                                = 125 J (d) time taken by the ball before reaches the ground T = 2usin/g    = 2*(50 m/s)(sin300) / 9.8 m/sec2    = 5.10 sec NOTE: YOU SHOULD NOT POST MULTIPLE QUESTIONS.                               = 1/2 (0.2 kg)(6.5 m/sec)                                = 0.65 J (c) Potential energy u = mgh    h is the maximum height = u2sin2/2g                                = (0.2 kg)(9.8 m/sec2)[(50m/s)2(sin300)2/2*9.8 m/sec2]                                = 125 J (d) time taken by the ball before reaches the ground T = 2usin/g    = 2*(50 m/s)(sin300) / 9.8 m/sec2    = 5.10 sec NOTE: YOU SHOULD NOT POST MULTIPLE QUESTIONS. time taken by the ball before reaches the ground T = 2usin/g    = 2*(50 m/s)(sin300) / 9.8 m/sec2    = 5.10 sec NOTE: YOU SHOULD NOT POST MULTIPLE QUESTIONS.

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