Physics

This is because the satellite doesn't have any other external third force acting upon it. Its original total energy comprises kinetic and potential energy, which remains constant since the gravity has zero influence over the satellite.

$g_{eff}=g+a$

$=g+\frac{g}{6}$

$=\frac{7g}{6}$

$T\text{'}=2\pi \sqrt[]{\frac{\mathcal{l}}{g_{eff}}}$

$T\text{'}=\sqrt[]{\frac{6}{7}}T$

According to question, we can write

  $d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2*125=500m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

According to question, we can write

  $d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2*125=500m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

Steam point (T₁) = 100°C = 273 + 100 = 373 K

Ice point (T₂) = 0°C = 273 K

$\eta =\left(1-\frac{T_2}{T_1}\right)*100=\left(1-\frac{273}{373}\right)*100=26.81\mathrm{\%}$  

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_P=\frac{{\mu }_{0}I}{2\pi r}*2$

$\Rightarrow 300*10^{-6}=\frac{4\pi *10^{-7}}{2\pi *4*10^{-2}}*2$                                                                         

=> I = 30 A

$g\propto r0\le r\le R$

$g\propto \frac{1}{r^2}r>R$

The minus sign in the formula indicates that:

• A gravitational force exists.
• The satellite cannot leave the orbit by itself.
• A minimum threshold of energy will be required to move the satellite from its existing orbit.

By conservation of Angular momentum

Li = Lf 

$M{R}^2\omega =\left(m{R}^2+2m{R}^2\right)\omega \text{'}$

$\frac{2M}{M+2m}=\omega \text{'}$

$\Rightarrow$ $\omega \text{'}=\frac{2M}{M+2m}$

Process-AB $\Rightarrow$ Isobaric,                                                           

Process-AC $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃