Physics Motion in Plane

A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously.

y = x5 (1 – x) = x tan θ$\left(1-\frac{x}{R}\right)$

tan = 5, R = 1

$sin\theta =\frac{5}{\sqrt[]{26}},cos\theta =\frac{1}{\sqrt[]{26}}$

$R=\frac{u^{2}sin2\theta }{g}=1$

$\Rightarrow u^2=26\Rightarrow u=\sqrt[]{26}m/s$

y – component of initial velocity

= u sin θ

= $=\sqrt[]{26}*\frac{5}{\sqrt[]{26}}$

= 5 m/s

H₁ = H₂

$\frac{u_1^{2}si{n}^2{\theta }_1}{2g}=\frac{u_2^{2}si{n}^2{\theta }_2}{2g}$

$\Rightarrow u_1^2{\left(sin30°\right)}^2={u_2}^2{\left(sin45°\right)}^2$

$\Rightarrow$$\frac{u_1}{u_2}$$=2$

For climbing downward

50 g – T = 50a

T = 500 – 50 * 4

= 300 N < 350 N

for climbing upwards

T – 50 g = 50 a

T – 500 = 50 * 5

T = 750 N > 350 N

Component of $\overrightarrow{A}an\overrightarrow{B}$ $\stackrel{}{}\stackrel{}{}$

$=\left(\overrightarrow{A}.\widehat{B}\right)\widehat{B}$

$\left[\left(\widehat{i}+\widehat{j}+\widehat{k}\right).\frac{\left(\widehat{i}+\widehat{j}\right)}{\sqrt[]{2}}\right]\frac{\left(\widehat{i}+\widehat{j}\right)}{\sqrt[]{2}}$

$=\frac{1}{2}\left(1+1\right)\left(\widehat{i}+\widehat{j}\right)$

$=\widehat{i}+\widehat{j}$

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

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

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

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

$\overrightarrow{A}.\overrightarrow{B}=\left|\overrightarrow{A}*\overrightarrow{B}\right|\Rightarrow ABcos\theta =ABsin\theta \Rightarrow \theta =45°$             

$\left|\overrightarrow{A}-\overrightarrow{B}\right|=\sqrt[]{A^2+B^2-2ABcos\theta }=\sqrt[]{A^2+B^2-2AB*\frac{1}{\sqrt[]{2}}}=\sqrt[]{A^2+B^2-\sqrt[]{2}AB}$                

$\left|\overrightarrow{P}\right|=\left|\overrightarrow{Q}\right|=x$

let the angle between $\overrightarrow{P}and\overrightarrow{Q}$ $\stackrel{}{}\stackrel{}{}$ is , so according to question, we can write $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\text{exception 25:}\text{exception 25:}$

^{$\left|\overrightarrow{P}+\overrightarrow{Q}\right|=n*\left|\overrightarrow{P}-\overrightarrow{Q}\right|\Rightarrow \sqrt[]{P^2+Q^2+2PQcos\theta }=n*\sqrt[]{P^2+Q^2-2PQcos\theta }$}

$\Rightarrow \sqrt[]{x^2+x^2+2x^{2}cos\theta }=n*\sqrt[]{x^2+x^2-2x^{2}cos\theta }$

$\Rightarrow \theta =co{s}^{-1}\left(\frac{n^2-1}{n^2+1}\right)$

Comparing E = 20 cos (2 * 10¹⁰ t – 200x) V/m to

$E=E_{0}cos\left(\omega t-kx\right)v/m$              

$\omega =2*10^{10},K=200$               

Speed = $\frac{2*10^{10}}{200}=10^{8}m/s$  

R.I. $=\frac{C}{speed}=\frac{3*10^8}{10^8}=3$  

$NowR.I.=\sqrt[]{{\epsilon }_r{\mu }_r}$

$3=\sqrt[]{{\epsilon }_r*1}$

${\epsilon }_r=9$              

Range R = $\frac{u^{2}sin2\theta }{g}$

For 42° and 48° Range will be same

$H_{max}\alpha$  maximum q

So maximum height will be for 48°