{Impulse} is the product of force and the time during which the force acts. It is equal to the change in momentum of an object. Mathematically, \( J = F \cdot t = \Delta p \).
Impulse (\( J \)) is defined as the change in momentum of an object. It can be calculated using the formula: \[ J = \Delta p = m \cdot v \] where:
\( m \) is the mass of the bullet,
\( v \) is the velocity of the bullet.
Given: \[ m = 10 \, \text{g} = 0.01 \, \text{kg} \\ v = 600 \, \text{m/s} \] Substituting the values: \[ J = 0.01 \, \text{kg} \times 600 \, \text{m/s} = 6 \, \text{Ns} \] Therefore, the impulse supplied to the gun is \( 6 \, \text{Ns} \).
A body of mass \( (5 \pm 0.5) \, \text{kg} \) is moving with a velocity of \( (20 \pm 0.4) \, \text{m/s} \). Its kinetic energy will be:
A solid sphere of mass $1 \,kg$ rolls without slipping on a plane surface Its kinetic energy is $7 \times 10^{-3} J$. The speed of the centre of mass of the sphere is ___$cm s ^{-1}$.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32