The point of contact of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2-2x+2y+1=0$ is

$A$ particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude.

In the following arrangement,$y = 1.0 \, mm$,$d = 0.24 \, mm$,and $D = 1.2 \, m$. The work function of the material of the emitter is $2.2 \, eV$. The stopping potential $V$ needed to stop the photocurrent will be....$V$

Under the influence of a uniform magnetic field,a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

$A$ gas is compressed at a constant pressure of $50 \ N/m^2$ from a volume of $10 \ m^3$ to a volume of $4 \ m^3$. Energy of $100 \ J$ is then added to the gas by heating. Its internal energy is:

Two particles of masses $m_1$ and $m_2$ in projectile motion have velocities $\vec{v}_1$ and $\vec{v}_2$ respectively at time $t = 0$. They collide at time $t_0$. Their velocities become $\vec{v}_1'$ and $\vec{v}_2'$ at time $2t_0$ while still moving in air. The value of $|(m_1 \vec{v}_1' + m_2 \vec{v}_2') - (m_1 \vec{v}_1 + m_2 \vec{v}_2)|$ is