Question
A body rotating at 20rad/s is acted upon by a constant torque providing it a deceleration of 2rad/s2. At what time will the body have kinetic energy same as the initial value if the torque continues to act?
$\Rightarrow\text{t}_1=\frac{\omega_2}{\alpha_1}=\frac{20}{2}=10\text{sec}$
Therefore 10sec it will come to rest. Since the same torque is continues to act on the body it will produce same angular acceleration and since the initial kinetic energy = the kinetic energy at a instant. So initial angular velocity = angular velocity at that instant Therefore time require to come to that angular velocity,$\Rightarrow\text{t}_2=\frac{\omega}{\alpha_2}=\frac{20}{2}=10\text{sec}$
therefore time required = t1 + t2 = 20sec.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\text{P}=\text{a}^3\text{b}^3/\big(\sqrt{\text{c}}\text{ d}\big)$
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?Calculate: