Physics

$R=\frac{\Delta V}{\Delta l}=\frac{0.75-0.70}{\left(5-3\right)*10^{-3}}=\frac{0.05*1000}{2}=25\Omega$

${?}^2={?}_0^2+2???{\left(\frac{2?}{60}*1800\right)}^2={\left(\frac{2?}{60}*600\right)}^2+2\left(\frac{2?}{60}*\frac{1200}{10}\right)?$

$?{\left(60?\right)}^2?{\left(20?\right)}^2=2\left(4?\right)???=\frac{80?*40?}{2*4?}=400?\text{?}\text{?}rad$

Number of rotations = $\frac{?}{2?}=\frac{400?}{2?}=200$

$l=JAcos\theta \Rightarrow 5=J*0.04*cos60°\Rightarrow J=\frac{5}{0.02}=250A/m^2$

$J=\frac{E}{\rho }\Rightarrow E=\rho J=44*10^{-8}*250=11*10^{-5}V/m$              

Power dissipated in cycle = P = $\frac{V_0{l}_0}{2}cos?=\frac{V_0^2}{2z}*\frac{R}{z}=\frac{V_0^{2}R}{2z}$  

For resonance, z = R, so

$\Rightarrow P=\frac{V_0^2}{2R}=\frac{250\sqrt[]{2}*250\sqrt[]{2}}{2*5}=12500Watt=125*10^{2}Watt$        

Consider translational motion of a body

mg sinq - f_S = ma             .(i)

Consider rotational motion of body about its centre of mass

$f_{S}R=l\alpha \Rightarrow \alpha =\frac{f_{S}R}{l}$ .(ii)

Using the condition of rolling without slipping at contact, we have

$\alpha R=a\Rightarrow \frac{f_S{R}^2}{l}=\frac{mgsin\theta -f_S}{m}\Rightarrow f_S=\frac{lmgsin\theta }{l+m{R}^2}$

Using the condition of static friction, we have

$f_S\le \mu N\Rightarrow \frac{lmgsin\theta }{l+m{R}^2}\le \mu mgcos\theta \Rightarrow \mu \ge \frac{ltan\theta }{l+m{R}^2}...........\left(iii\right)$

$\alpha =\frac{mgRsin\theta }{l+m{R}^2}\Rightarrow a=\alpha R=\frac{mg{R}^{2}sin\theta }{l+m{R}^2}=\frac{g{R}^{2}sin\theta }{k^2+R^2}...........\left(iv\right)$

$t=\sqrt[]{\frac{2\mathcal{l}}{a}}\Rightarrow$ Time required to reach the ground

$\Rightarrow \frac{v_{ring}}{v_{cylinder}}=\sqrt[]{\frac{2g\mathcal{l}{R}^{2}sin\theta }{k_{ring}^2+R^2}}*\sqrt[]{\frac{k_{cylinder}^2}{2g\mathcal{l}{R}^{2}sin\theta }}=\sqrt[]{\frac{\frac{R^2}{2}+R^2}{R^2+R^2}}=\frac{\sqrt[]{3}}{2}$

N_a -> Avogadro's Number

$J{\beta }^2\equiv \left[\mu kR\right]\Rightarrow {\beta }^2\equiv \left[kR\right]\Rightarrow \beta \equiv \left[k\right]\equiv \left[M{L}^2{T}^{-2}{K}^{-1}\right]$

, and

-> $S\equiv {\alpha }^2\beta \Rightarrow \left[M{L}^2{T}^{-2}{K}^{-1}\right]\equiv {\alpha }^2\left[M{L}^2{T}^{-2}{K}^{-1}\right]\Rightarrow \alpha \equiv \left[M^0{L}^2{T}^0{K}^0\right]$ Dimensionless

Let the work function of metal is $?$ , so

$k_1=3V_{0}e=\frac{hc}{\lambda }-?$ . (i)

$K_2=V_{0}e=\frac{hc}{2\lambda }-?$ . (ii)

Doing (i) – 3 X (ii), we have

$0=\frac{hc}{\lambda }-\frac{3hc}{2\lambda }+2?\Rightarrow ?=\frac{hc}{4\lambda }$    

The threshold wavelength = $\frac{hc}{?}=4\lambda$

$A=-\frac{dN}{dt}=\lambda N=\lambda N_0{e}^{-\lambda t}\Rightarrow \lambda \frac{N_0}{16}=\lambda N_0{e}^{-80\lambda }\Rightarrow \frac{1}{16}=e^{-80\lambda }\Rightarrow \frac{1}{16}=e^{-80\lambda }$

$\Rightarrow 80\lambda =\mathcal{l}n\left(16\right)=4\mathcal{l}n\left(2\right)\Rightarrow t_{\frac{1}{2}}=\frac{\mathcal{l}n\left(2\right)}{\lambda }=\frac{80}{4}=20days$

This is because acceleration depends on both force and mass (F = ma).

We know that from Newton's Third Law. While action-reaction forces are always equal, the objects they act on usually have very different masses.

If you consider a falling stone and the Earth as an action-reaction pair. The Third Law tells us,  

• Earth pulls stone down with force F

• Stone pulls Earth up with equal force F

• But Earth's mass is enormous, so its acceleration is tiny (F/huge mass = tiny acceleration)

• Stone's mass is small, so its acceleration is large (F/small mass = large acceleration)

• Result: Stone falls noticeably, Earth's motion is unnoticeable.

The third law of motion by Newton may confuse you into thinking that action causes reaction in sequence. This is incorrect.

In reality, both forces exist at the exact same moment. When you push a wall, your hand pushes the wall. At the same time, the wall pushes back simultaneously. Not one, then the other. Just note that there's no time delay between them.