(8) Going up, the ball will experience the force due to gravity (mg) and the force due to air resistance (F) added together, so the magnitude of the acceleration (force divided by mass) will be g + F/m. On the way down, it will experience the force due to gravity pulling down on it and the force due to air resistance pulling up on it, so the magnitude of the acceleration will be g + F/m. (C)

(15) Potential energy = mgh. Work done = work done against air pressure above gas + work done to lift mass sitting on top. = pAh + mgh.
So the ratio is mg/(pA + mg). (B)

(16) Trying to get my head round this one still!

(37) Total voltage 4.1V + 3.7V should equal the total voltage 2.2V + 1.2V + V_R (where V_R is the voltage across R) as they are the two branches in the parallel part of the circuit.
So 7.8V = 3.4V + V_R , so V_R = 4.4V (A)

Reply 5

Teenie2

11

(16) The system has lost its kinetic energy (to friction with the air). KE = 0.5 ky² where y is the amplitude of the motion and k is the spring constant. At rest, the forces are balanced and mg = kx where x is the distance between the position of the unstreched spring and the new equilibrium point, so this gives us a value for k = mg/x = 8/0.02 = 400.
Back to KE = 0.5 ky² . The amplitude y = 0.02 so KE = 0.5 * 400 * 0.02² = 0.08J ( B)

Reply 6

HUSSAI55

OP

10

thanks alot!! :smile:

Original post by Teenie2

(16) The system has lost its kinetic energy (to friction with the air). KE = 0.5 ky² where y is the amplitude of the motion and k is the spring constant. At rest, the forces are balanced and mg = kx where x is the distance between the position of the unstreched spring and the new equilibrium point, so this gives us a value for k = mg/x = 8/0.02 = 400.
Back to KE = 0.5 ky² . The amplitude y = 0.02 so KE = 0.5 * 400 * 0.02² = 0.08J ( B)

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