An elevator starts from rest with a constant downward acceleration and covers `2.5m` in first second If the lift weighs `200kg` what would be the tension in the ropes of the lift ?

A lift starts from rest with a constant upward acceleration It moves 1.5 m in the first 0.4 A person standing in the lift holds a packet of 2 kg by a string Calculate the tension in the string during the motion .

A lift starts from rest with a constant upward acceleration It moves 1.5 m in the first 0.4 A person standing in the lift holds a packet of 2 kg by a string Calculate the tension in the string during the motion .

An elevator starts from rest with a constant upward acceleration it moves 2m in the first 0.6 second. A passenger in the elevator is holding a 3 kg package by a vertical string When the elevator is moving What is the tension in the string?

An elevator starts from rest with a constant upward acceleration. It moves 5 m in the first 2 s. A passenger in the elevator is holding a 4 kg package by a vertical string. Find the tension in the string during the acceleration process in N.

A lift starts from rest with constant upward acceleration . It moves 1.5m in the first 0.4 secs. A person standing in the lift hold a packet of 2 kg by a string. Calculate the tension in the string during the motion.

A body ,starting from rest and moving with constant acceleration,covers 10m in the first second.Find the distance travelled by it in the next second.

A body ,starting from rest and moving with constant acceleration,covers 10m in the first second.Find the distance travelled by it in the next second.

A car starts from rest and moves with constant acceleration. In first t seconds it covers distance x. Then distance covered by it in next t seconds will be:-

A car starts from rest and moves with constant acceleration. The ratio of distance covered by the car in nth second with that covered in n seconds is: