Physics

${\mathrm{h}}_{\text{max}}=\frac{{\mathrm{u}}^2{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta }{2\mathrm{g}}=\frac{280*280}{2*9.8}*\frac{1}{4}$

=1000 m

Initial velocity $=-v\hat{j}$

Final velocity $=-v\hat{j}$  

Change in velocity $=v\hat{i}-(-v\hat{j})$

$=v(\hat{i}+\hat{j})$ Momentum gain is along $\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}$

$\Rightarrow$ Force experienced is along $\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}$

$\Rightarrow$ Force experienced is in North-East direction.

Average speed $=\left(\frac{4{\mathrm{v}}^2}{3\mathrm{v}}\right)$

$=\frac{4\mathrm{v}}{3}$

$m=\rho \pi r^{2}1$

$\begin{array}{rr}& \rho =\frac{\mathrm{m}}{\pi {\mathrm{r}}^{2}1}\\ & \frac{\mathrm{\Delta }\rho }{\rho }=\frac{\mathrm{\Delta }\mathrm{m}}{\mathrm{m}}+\frac{2\mathrm{\Delta }\mathrm{r}}{\mathrm{r}}+\frac{\mathrm{\Delta }\mathrm{l}}{1}\\ & \frac{\mathrm{\Delta }\rho }{\rho }*100=\frac{0.002}{0.4}*100+\frac{2*0.001}{0.3}*100+\frac{0.02}{5}*100\\ & =\frac{0.2}{0.4}+\frac{0.2}{0.3}+\frac{2}{5}\\ & =0.5+0.67+0.4\\ & =1.57\\ & =1.6\mathrm{\%}\end{array}$

As the factors controlling temperature and voltage supply are beyond prediction and control so the error occurred due to unpredictable fluctuations of temperature and voltage would be random errors.

A reverse-biased Zener diode is used as a voltage regulator.
The potential barrier for Germanium (Ge) is approximately 0.3 V.
The potential barrier for Silicon (Si) is approximately 0.7 V.

Yes, an object moving in a straight path can also possess angular momentum with respect to a referenc epoint which doesn't lie on it's path. In such cases. the referenc epoint may lie perpendicular distance from the axis. If an object is moving in a straight line (not through the reference point) the perpendicular distance $r_\perp$ (l = rp) will be non-zero.

Yes, how much time these planets take in completing an orbit around the sun hugely depends on their distance from the sun as well as their mass. Since these planets move in a circular orbit, the term angular momentum is used in such cases for computing relevant information.