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Answer 1

$\large\underline{\pink{\sf \orange{\bigstar} Given:-}}$⠀⠀⠀⠀⠀⠀⠀⠀

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The masses of four particles are

First particle (m₁) = m

Second particle (m₂) = 2m

Third particle (m₃) = 3m

Fourth particle (m₄) = 4m

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$\large\underline{\red{\sf \orange{\bigstar} To Find:-}}$

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The centre of mass of the system about the mass m placed at the origin.

The co-ordinated in which the particles are lying are,

→ (0,0) ; (1,0) ; (1,1) ; (0,1)

$\boxed{\boxed{\bf x_1 = 0,\ x_2 = 1,\ x_3 = 1,\ x_4 = 0 }}$

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$\boxed{\boxed{\bf y_1 = 0,\ y_2 = 0,\ y_3 = 1,\ y_4 = 1 }}$

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$\large\underline{\green{\sf \orange{\bigstar} Now:-}}$

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Applying for the formula for the X-component of the centre of mass,

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$\boxed{X_{cm} = \dfrac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}+m_{4}x_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}}$

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$\large\underline{\blue{\sf \orange{\bigstar} Putting \:\:the\:\: respective\:\: values\:\: of,\:\: we\:\: get }}$⠀⠀⠀⠀⠀

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$X_{cm} = \dfrac{m(0)+2m(1)+3m(1)+4m(0)}{m+2m+3m+4m}$

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$X_{cm} = \dfrac{2m+3m}{10m}$

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$X_{cm} = \dfrac{5\not{m}}{10 \not{m}}$

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Both numerator and denominator can be divided by 2.

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$\boxed{\boxed{X_{cm} = \dfrac{1}{2}}}$

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Applying for the formula for the Y-component of the centre of mass,

$\boxed{Y_{cm} = \dfrac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}+m_{4}y_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}}$

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$\large\underline{\red{\sf \orange{\bigstar} Putting \:\:the\:\: respective\:\: values\:\: of,\:\: we\:\: get }}$

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$Y_{cm} = \dfrac{m(0)+2m(0)+3m(1)+4m(1)}{m+2m+3m+4m}$

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$Y_{cm} = \dfrac{3m+4m}{10m}$

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$X_{cm} = \dfrac{7\not{m}}{10\not{m}}$

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$X_{cm} = \dfrac{7}{10}$

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$\large\underline{\green{\sf \orange{\bigstar} Hence:-}}$

The coordinates for the centre of mass should be written as,

$\boxed{(X_{cm}, Y_{cm})=(\frac{1}{2},\frac{7}{10})}$⠀⠀⠀⠀⠀⠀⠀⠀

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$\large\underline{\red{\sf \orange{\bigstar} Note:-}}$⠀⠀⠀⠀⠀

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$\large\underline{\purple{\sf \orange{\bigstar} Diagram\:\: is\:\: in\:\: attachment! }}$

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given Equation:-

⠀⠀⠀⠀ $\sf{\bigg[ \dfrac{5 {x}^{2} - 10}{12 } \bigg] }$

To find:-

value of x

Solution:-

use factor theorem

take the value of equation =0

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow \left [\dfrac {5x^2-10}{12}\right]=0$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow \dfrac {5x^2-10}{12}=0$

using cross multiplication

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow 5x^2-10=0$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow 5x^2=10$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x^2=\dfrac {10}{5}$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x^2=2$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x=\sqrt {2}$

$\\\\\therefore\sf x=\sqrt {2}.$