The energy required to accelerate a car from | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $10\, m/s$

Similar Questions

A ball moving with velocity $2\, m/s$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is $0.5$, then their velocities (in $m/s$) after collision will be

A light and a heavy body have equal kinetic energy. Which one has a greater momentum

The potential energy of a diatomic molecule is given by $U = \frac{A}{{{r^{12}}}} - \frac{B}{{{r^6}}}$ . $A$ and $B$ are positive constants. The distance $r$ between them at equilibrium is

The length of a spring is $\alpha $ when a force of $4\,N$ is applied on it and the length is $\beta $ when $5\,N$ force is applied. Then the length of spring when $9\,N$ force is applied is

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to