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Question

(d) Kinetic energy for first condition

=$\frac{1}{2}m\left( {v_2^2 - v_1^2} \right) = \frac{1}{2}m\left( {{{20}^2} - {{10}^2}} \right)$= $150\;mJ$

Vedclass · Accepted Answer

$3$ times

Vedclass · Answer

(d) Kinetic energy for first condition

=$\frac{1}{2}m\left( {v_2^2 - v_1^2} ight) = \frac{1}{2}m\left( {{{20}^2} - {{10}^2}} ight)$= $150\;mJ$

$K.E.$ for second condition = $\frac{1}{2}m\left( {{{10}^2} - {0^2}} ight) = 50mJ$

$	herefore \frac{{(K.E.)_I}}{{(K.E.)_{II}}} = \frac{{150m}}{{50m}} = 3$

The energy required to accelerate a car from $10 \,m/s$ to $20\, m/s$ is how many times the energy required to accelerate the car from rest to $10\, m/s$

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