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

The bob of a pendulum of length $l$ is pulled aside from its equilibrium position through an angle $\theta $ and then released. The bob will then pass through its equilibrium position with speed $v$ , where $v$ equals

A body of mass $m$ is moving in a circle of radius $r$ with a constant speed $v$. The force on the body is $\frac{{m{v^2}}}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle

Assume the aerodynamic drag force on a car is proportional to its speed. If the power output from the engine is doubled, then the maximum speed of the car.

If $F = 2x^2 -3x -2$, then choose correct option