A simple numerical on momentum
Car mass is 1200 kg moving velocity is 108 km/hr. On applying brakes velocity reduced to 36 km/hr. Find change in momentum? (posted by Aditi) Answer: m=1200 kg u=108 x 5/18=30m/s v=36×5/18=10m/s Change in momentum = m(v-u) = 1200 x (30-10)=1200 ...
Categories: Ask Physics, brakes, Force, Mechanics, Project Tags: 10m, Aditi, answer, brakes, car, change, mass, momentum, Physics, post, search terms, speed, velocity
A simple numerical on momentum
Car mass is 1200 kg moving velocity is 108 km/hr. On applying brakes velocity reduced to 36 km/hr. Find change in momentum? (posted by Aditi) Answer: m=1200 kg u=108 x 5/18=30m/s v=36×5/18=10m/s Change in momentum = m(v-u) = 1200 x (30-10)=1200 ...
Categories: Ask Physics, brakes, Force, Mechanics, Project Tags: 10m, Aditi, answer, brakes, car, change, link, mass, momentum, Physics, post, speed, velocity
Numerical Problem from Friction
“The friction coefficient between a road and the tyre of a vehicle is 4/3.find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding,the vehicle going down at a speed of 10m/s is stopped within 5m”...
Categories: Ask Physics, banking, brakes, Force, friction coefficient, friction forces, Mechanics, Project, roads, skidding, tyre, wheel Tags: 10m, brakes, coefficient, friction, friction coefficient, numerical, numerical problem, Physics, post, speed, tyre, vehicle, wheel
