Question

a force of 200 dyne acts on a body of mass 10 g for 5 sec. what will be the final velocity of the body if it starts from rest​

Answer 1

Answer:

1m/s

Explanation:

accl.=F/m

=20cm/s²

=0.2m/s²

V=u+at

where t=5 sec

V=0+5×0.2

= 1m/s

Answer 2

Given,

Given,Force, F = 200 dyne = 200*10^-5 = 0.002 N

Given,Force, F = 200 dyne = 200*10^-5 = 0.002 NMass, m = 10 g = 0.01 kg

Given,Force, F = 200 dyne = 200*10^-5 = 0.002 NMass, m = 10 g = 0.01 kgAcceleration, a = F/m = 0.2 m/s^2

Given,Force, F = 200 dyne = 200*10^-5 = 0.002 NMass, m = 10 g = 0.01 kgAcceleration, a = F/m = 0.2 m/s^2Initial velocity, u = 0 m/s

Given,Force, F = 200 dyne = 200*10^-5 = 0.002 NMass, m = 10 g = 0.01 kgAcceleration, a = F/m = 0.2 m/s^2Initial velocity, u = 0 m/sTime, t = 5 s

Given,Force, F = 200 dyne = 200*10^-5 = 0.002 NMass, m = 10 g = 0.01 kgAcceleration, a = F/m = 0.2 m/s^2Initial velocity, u = 0 m/sTime, t = 5 sFinal velocity, v = u + at = 0 + 0.2*5 = 1 m/s

Hope it is helpful.