Class 11th

According to Newton's law of motion, we can write

ma = mg – N = mg - $\frac{mg}{4}=\frac{3mg}{4}$

$\Rightarrow a=\frac{3g}{4}$                             

According to conservation of energy, we can write

Gain in kinetic energy = Loss in potential energy

=>K_f – K_in = U_in - U_f

=>K - = mgy – mg (y – y₀) = mgy₀

$Mean=\frac{3+12+7+a+\left(43-a\right)}{5}=13$  

Variance =  $\frac{3^2+12^2+7^2+a^2+{\left(43-a\right)}^2}{5}-{\left(13\right)}^2$  

$\Rightarrow \frac{2a^2-a+1}{5}\to Naturalnumber$  

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not  $4\lambda or4\lambda +1from.$  

As each square form is  $4\lambda or4\lambda +1$  

According to Newton's laws of motion, we can write

4a = 4g – T . (1), and

40a = T - f_k = T - $\mu$ (40g) ……………… (2)

Adding equations (1) and (2), we can write

44a = 40 – 0.02 * (400) = 32

$\Rightarrow a=\frac{32}{44}=\frac{8}{11}m/s^2$  

Total number of possible relation = $2^{n^2}=2^4=16$  

Favourable relations = $?,\left\{\left(x,x\right)\right\},\left\{\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right),\left(x,y\right),\left(y,x\right)\right\}$

Probability =  $\frac{5}{16}$  

At each point in the orbit, there is a variation in the kinetic and potential energy of the satellite. If one increases, the other one decreases. Similarly, if the other one increases, the first one decreases. But their amount of increment or decrement is such that the total sum i.e. mechanical energy of the satellite remains the same. This in turn allows free movement.

According to question, we can write

Total moles of gas = n = n_Oxygen + n_Oxygen = $\frac{16}{2}+\frac{128}{32}=12moles$  

Volume of gas = 12 * 22.4 litre = 268.8 litre = 2.688 * 10⁵ cm³ $\approx 27*10^{4}c{m}^3$  

$KE=\Delta U$

$\Delta U=\frac{1}{2}m{v}^2$

$n{C}_v\Delta T=\frac{m{v}^2}{2}$

$\Rightarrow \frac{m}{M}\frac{R}{y-1}\Delta T=\frac{m{v}^2}{2}$

$\Delta T=\frac{0.4}{2}\frac{M{v}^2}{R}$

$\Delta T=\frac{M{v}^2}{5R}$

According to question, we can write

$\omega =\pi =\sqrt[]{\frac{g}{\mathcal{l}}}\Rightarrow \mathcal{l}=\frac{g}{{\pi }^2}=\frac{9.8}{{\left(3.14\right)}^2}=0.99395=99.4cm$

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.