The switch in the circuit in the figure is in position $P$ for a long time and then moved to position $Q$ at time $t=0$

The value of $\frac{d v(t)}{d t}$ at $t=0^{+}$is

In the circuit shown in the figure, the switch is closed at time $t=0$, while the capacitor is initially charges to -5 V (i.e., $\left(\theta_C(0)=-5 \mathrm{~V}\right)$.

The time after which the voltage across the capacitor becomes zero (rounded off to three decimal places) is $\_\_\_\_$ ms .

Consider a real-valued base-band signal $x(t)$. band limited to 10 kHz . The Nyquist rate for the signal $y(t)=x(t) \times \left(1+\frac{t}{2}\right)$ is