Two masses 90 kg and 160 kg are separated b | vedclass

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(B) Let the two masses be $m_1 = 90 \ kg$ at point $A$ and $m_2 = 160 \ kg$ at point $B$. The point $C$ is at a distance $r_1 = 3 \ m$ from $A$ and $r

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$9.43 	imes 10^{-10} \ N \ kg^{-1}$

Vedclass · Answer

(B) Let the two masses be $m_1 = 90 \ kg$ at point $A$ and $m_2 = 160 \ kg$ at point $B$. The point $C$ is at a distance $r_1 = 3 \ m$ from $A$ and $r_2 = 4 \ m$ from $B$. The distance between $A$ and $B$ is $5 \ m$.Since $3^2 + 4^2 = 5^2$,the triangle $ABC$ is a right-angled triangle with the right angle at $C$.The gravitational field intensity at $C$ due to mass $m_1$ is $E_A = \frac{G m_1}{r_1^2} = \frac{G 	imes 90}{3^2} = 10G$.The gravitational field intensity at $C$ due to mass $m_2$ is $E_B = \frac{G m_2}{r_2^2} = \frac{G 	imes 160}{4^2} = 10G$.Since the angle between $E_A$ and $E_B$ is $90^\circ$,the resultant intensity $E$ is given by:$E = \sqrt{E_A^2 + E_B^2} = \sqrt{(10G)^2 + (10G)^2} = 10G\sqrt{2}$.Substituting $G = 6.67 	imes 10^{-11} \ N \ m^2 \ kg^{-2}$:$E = 10 	imes 6.67 	imes 10^{-11} 	imes 1.414 \approx 9.43 	imes 10^{-10} \ N \ kg^{-1}$.

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