A body weighs 49 , N on a spring balance at t | vedclass

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(B) The weight of the body at the poles is given by $W_p = mg = 49 \, N$. Since $g = 9.8 \, m/s^2$,the mass of the body is $m = \frac{49}{9.8} = 5 \,

Vedclass · Accepted Answer

$48.83$

Vedclass · Answer

(B) The weight of the body at the poles is given by $W_p = mg = 49 \, N$. Since $g = 9.8 \, m/s^2$,the mass of the body is $m = \frac{49}{9.8} = 5 \, kg$. At the equator,the effective acceleration due to gravity is $g_e = g - R\omega^2$,where $\omega$ is the angular velocity of the Earth. The angular velocity $\omega = \frac{2\pi}{T}$,where $T = 24 	imes 3600 \, s$. $R\omega^2 = R \left(\frac{2\pi}{T}ight)^2 = 6.4 	imes 10^6 	imes \left(\frac{2 	imes 3.14}{86400}ight)^2 \approx 0.0337 \, m/s^2$. The weight at the equator is $W_e = m(g - R\omega^2) = 5 	imes (9.8 - 0.0337) = 5 	imes 9.7663 = 48.8315 \, N$. Rounding to two decimal places,the weight is $48.83 \, N$.

$A$ body weighs $49 \, N$ on a spring balance at the North Pole. What will be its weight recorded on the same weighing machine if it is shifted to the equator (in $, N$)? (Use $g = \frac{GM}{R^2} = 9.8 \, m/s^2$ and radius of the Earth,$R = 6400 \, km$.)

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