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
Answers
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
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.