Inside a horizontally moving box,an experimen | vedclass

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(B) The acceleration of the box is $a = 10\,m/s^2$. In the non-inertial frame of the box,a pseudo force $F_p = ma$ acts on the body in the direction o

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$10\sqrt{2}$

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(B) The acceleration of the box is $a = 10\,m/s^2$. In the non-inertial frame of the box,a pseudo force $F_p = ma$ acts on the body in the direction opposite to the acceleration of the box. Here,$m = 1\,kg$ and $a = 10\,m/s^2$,so the pseudo force $F_p = 1 	imes 10 = 10\,N$ acts horizontally. The gravitational force $mg = 1 	imes 10 = 10\,N$ acts vertically downwards. In the equilibrium position relative to the experimenter,the tension $T$ in the string must balance the resultant of the gravitational force and the pseudo force. $T = \sqrt{(mg)^2 + (ma)^2} = \sqrt{10^2 + 10^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2}\,N$.

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If a body of mass $m$ is carried by a lift moving with an upward acceleration $a$,then the forces acting on the body are:
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