A ship of mass 3 × 10^7kg initially at rest, is pulled by a force of 5 × 10^4N through a distance of 3m.

(d) : Acceleration of block (a) = (Force applied)/(Total mass) or  a = P/(M + m) Therefore, Force on block = Mass of block x a = MP/(M + m).

The correct answer is (a) vA > vB   Explanation:  Let T be the torque and A and A' be the angular accelerations. A = T/IA and A' = T/IB. Clearly A >A'. In...

(a) the measured times are same (b) the measured speeds are same EXPLANATION:   The time of fall t = √(2h/g) and the final speed v=√(2gh). The time period of the pendulum clock T=2π√(l/g)....

(a) the measured times are same (b) the measured speeds are same EXPLANATION:   The time of fall t = √(2h/g) and the final speed v=√(2gh). The time period of the pendulum clock T=2π√(l/g). If the...

correct answer is (C) g.

(A) F1(1 -y/L)  Explanation: acceleration of string =F/M,  mass of portion of string between distance y and other end of string=M(1-y/L).  tension at distance y=M(1-y/L)F/M =F(1-y/L)

Retarding force, F = -50 N   Mass of the body, m= 20 kg   Initial velocity of the body, u= 15 m/s   Final velocity of the body, v= 0   UsingNewton's second law of motion, the...

 a = - 20ms-2  0 ≤ t ≤ 30s  t = - 5s:  x = ut = -10 × 5 = - 50m  t = 25s :  x = ut + (½) at2 = (10 × 25...

(b) Mate receipt