Physics

V = 10 $\left(1?e^{?t/RC}\right)$ ${}^{}$

2 = 20 $\left(1?e^{?t/RC}\right)$ ${}^{}$

$\frac{1}{10}=1?e^{?t/RC}$

$e^{t/RC}=\frac{10}{9}$

$\frac{t}{RC}=ln\left(\frac{10}{9}\right)=0.105$

$C=\frac{t}{R*0.105}=\frac{10^{?6}}{10*0.105}=0.95?F$

As observer is at O So height of water observed by observer

$\Rightarrow \frac{H}{{\mu }_{\omega }}=\frac{H}{\left(4/3\right)}=\frac{3H}{4}$

given diagram (17.5 H) is height of observer

$So\frac{3H}{4}=17.5-H$

$\frac{7H}{4}=17.5$

7H = 70

H = 10

It's the energy that is needed for a change of state to transition from one substance or physical state to another. Remember that the main condition for latent heat is that there is no temperature change when the energy is absorbed or released. 

You will be seeing these temperature plateaus appearing on the heating curve at the main melting and boiling points. The reason for that is the energy reaches latent heat. And we know that latent heat is the energy that lets phase transitions occur without increasing the temperature. 

$v=\frac{fu}{u-f}=\frac{(-10)(-15)}{-15+10}\mathrm{c}\mathrm{m}=-30\mathrm{c}\mathrm{m}$

$v_{\text{image}}=\frac{dv}{dt}=-{\left(\frac{v}{u}\right)}^2\frac{du}{dt}=-{\left(\frac{-30}{-15}\right)}^2\left(10\right)\mathrm{c}\mathrm{m}{\mathrm{s}}^{-1}=-40\mathrm{c}\mathrm{m}{\mathrm{s}}^{-1}$

$V_{\text{object}}=\frac{du}{dt}=+10\mathrm{c}\mathrm{m}{\mathrm{s}}^{-1}$

Relative velocity $=v_{\text{object}}-v_{\text{image}}$

$=10-(-40)=50\mathrm{c}\mathrm{m}{\mathrm{s}}^{-1}$

${\eta }_1=1-\frac{T_2}{T_1},{\eta }_2=1-\frac{T_3}{T_2}$

as ${\eta }_1={\eta }_2\Rightarrow \frac{T_2}{T_1}=\frac{T_3}{T_2}\Rightarrow T_2={\left(T_1{T}_3\right)}^{\frac{1}{2}}=\sqrt[]{400*900}=600\mathrm{K}$

$M={\chi }_{m}H$ and ${}_{}\frac{}{}$ ${\chi }_m=\frac{C}{T}($ Curie law $$ $)$ ,

$$ $M=\frac{CH}{T}$  or $\frac{M_1}{M_2}=\frac{C{H}_1/T_1}{C{H}_2/T_2}=\left(\frac{H_1}{H_2}\right)\left(\frac{T_2}{T_1}\right)$ $\frac{{}_{}}{{}_{}}\frac{{}_{}{}_{}}{{}_{}{}_{}}\left(\frac{{}_{}}{{}_{}}\right)\left(\frac{{}_{}}{{}_{}}\right)$

$M_2=M_1\left(\frac{H_2}{H_1}\right)\left(\frac{T_1}{T_2}\right)=12(\mathrm{A}/\mathrm{m})\left[\frac{0.2T}{0.6T}\right]\left[\frac{4\mathrm{K}}{16\mathrm{K}}\right]$

$=1\mathrm{A}/\mathrm{m}$

 Newton's law of cooling, $\begin{array}{rr}& -\frac{dT}{dt}=k\left(T-T_0\right)\Rightarrow -\int \frac{dT}{T-T_0}=k\int dt\\ & \Rightarrow \mathrm{}{\left[-\mathrm{l}\mathrm{n}?\left|T-T_0\right|\right]}_T^{\left(T+T_0\right)/2}=kt\mathrm{}\left(?\text{Heat}\propto \left(T-T_0\right)\right)\\ & \Rightarrow \mathrm{}-\mathrm{l}\mathrm{n}?\left|\frac{\left(T-T_0\right)/2}{T-T_0}\right|=kt\mathrm{}\Rightarrow \mathrm{}-\mathrm{l}\mathrm{n}?\left|\frac{1}{2}\right|=kt\mathrm{}\Rightarrow \mathrm{}t=\frac{\mathrm{l}\mathrm{n}?2}{k}=\frac{\mathrm{l}\mathrm{n}?4}{2k}\mathrm{}\therefore \mathrm{}n=4\end{array}$

Kindly consider the following Image 

Kindly consider the following Image