Physics

Zero velocity doesn't mean zero acceleration. When you throw a ball-like object upward, at its peak the velocity is zero. But acceleration remains constant due to gravity. The velocity is still changing from positive to negative at that instant. That means acceleration continues.

70 g – N = 70 a = 70 * 0.2 = 14

->N = 700 – 14 = 686 N

$F=\frac{q\left(Q-q\right)}{4\pi {\epsilon }_0{r}^2}\Rightarrow \frac{dF}{dq}=\frac{1}{4\pi {\epsilon }_0{r}^2}\left(Q-2q\right)\Rightarrow \frac{dF}{dq}=\frac{1}{2\pi {\epsilon }_0{r}^2}$

Here r is fixed

For maxima or minima of force, its first derivative should be zero.

$\frac{dF}{dq}=\frac{1}{4\pi {\epsilon }_0{r}^2}\left(Q-2q\right)=0\Rightarrow q=\frac{Q}{2}$                

Since second derivative is always negative so maxima will occur at this value of q.

$P_z=V_z{l}_z\Rightarrow 0.5=8l_z\Rightarrow l_z=l_P=\frac{1}{16}Amp$

When zener is connected across a potential divider arranged with maximum potential drop across zener diode, then

$$ $V_P=V-V_z=20-8=12$  volt Potential difference across protective resistance RP

$R_P=\frac{V_P}{I_P}=\frac{12}{\frac{1}{6}}=192Amp.$

For ascent

$t_a=\sqrt[]{\frac{2\mathcal{l}}{a_{ascent}}}=\sqrt[]{\frac{2\mathcal{l}}{g\left(sin\theta +\mu cos\theta \right)}}$

For descent

$t_d=\sqrt[]{\frac{2\mathcal{l}}{a_{descent}}}=\sqrt[]{\frac{2\mathcal{l}}{g\left(sin\theta -\mu cos\theta \right)}}$

According question, we can write

$\Rightarrow \mu =\frac{3}{5}tan\theta =\frac{3}{5}tan30°=\frac{\sqrt[]{3}}{5}$

${\overrightarrow{v}}_{BW}=4\sqrt[]{2}\left(cos45°\widehat{i}+sin45°\widehat{j}\right)=4\widehat{i}+4\widehat{j},and{\overrightarrow{v}}_{Wg}=0\widehat{i}-\widehat{j}$              

${\overrightarrow{v}}_{Bg}={\overrightarrow{v}}_{BW}+{\overrightarrow{v}}_{Wg}=\left(4\widehat{i}+4\widehat{j}\right)+\left(0\widehat{i}-\widehat{j}\right)=4\widehat{i}+3\widehat{j}$      

$\left|{\overrightarrow{v}}_{Bg}\right|=5m/s\Rightarrow SpeedofButterfly$

Magnitude of displacement of Butterfly = $\left|{\overrightarrow{v}}_{Bg}\right|*t=5*3=15m$       

Here

$R_{eq}=\left(12/{/}^{\mathcal{l}}6\right)+\left(4/{/}^{\mathcal{l}}4\right)+\left(6/{/}^{\mathcal{l}}12\right)$

$=\left(\frac{12*6}{18}\right)+\left(\frac{4*4}{8}\right)+\left(\frac{6*12}{18}\right)$

= 4 +2 + 4 = 10 $\Omega$

$C=\frac{{\epsilon }_{0}A}{d}\Rightarrow X_C=\frac{1}{C\omega }=\frac{d}{{\epsilon }_{0}A\omega }\Rightarrow l_0=\frac{V_0}{X_C}=\frac{V_0{\epsilon }_{0}A\omega }{d}=\frac{2\pi f{V}_0{\epsilon }_{0}A}{d}$

$\Rightarrow l_0=\frac{2*3.14*50*20*8.85*10^{-12}*1}{2*10^{-3}}=27.79*10^{-6}A=27.79\mu A$              

Since process is isothermal, so

$W=nRT\mathcal{l}n\left(\frac{V_f}{V_i}\right)=1*8.3*300*\mathcal{l}n\left(\frac{4}{2}\right)=1*8.3*300*\mathcal{l}n\left(2\right)$       

$\Rightarrow W=1*8.3*300*0.6931\approx 1725.82J\approx 17258*10^{-1}J$

Assuming Tension in metal wire suspended from roof varies linearly and 0 and T₀ ( developed due its own weight) are tensions are ends of wire of length $\mathcal{l}.$ So

$T_1\alpha \left({\mathcal{l}}_1-\mathcal{l}\right),and{T}_2\alpha \left({\mathcal{l}}_2-\mathcal{l}\right)$

$\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}\Rightarrow \mathcal{l}=\frac{T_1{\mathcal{l}}_2-T_2{\mathcal{l}}_1}{T_1-T_2}$