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:

$A$ wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$,its length increases by $l$. Another wire of the same material of length $2L$ and radius $2r$ is pulled by a force $2f$. Find the increase in length of this wire.

In $\triangle ABC$, the midpoints of the sides $AB, BC$, and $CA$ are respectively $(l, 0, 0), (0, m, 0)$, and $(0, 0, n)$. Then, $\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2}$ is equal to

$A$ wheel has an angular acceleration of $3.0\, rad/s^2$ and an initial angular speed of $2.00\, rad/s$. In a time of $2\, s$,it has rotated through an angle (in radians) of:

Let $a = \min \{x^2 + 2x + 3, x \in R\}$ and $b = \lim_{x \to 0} \frac{\sin x \cos x}{e^x - e^{-x}}$. Then the value of $\sum_{r=0}^n a^r b^{n-r}$ is

Along the $X$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at $x=a+r$ (if $a \ll r$) is: ($\varepsilon_0$ is the permittivity of free space)