The radius of the Earth is approximately $6.4 \times 10^{6} m$ . The instantaneous velocity of a point on the surface of the Earth at the equator is $4.672 \times 10^{x}$ . Find the value of x :

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Solution

Radius of earth $r = 6.4 \times 10^{6}$ m
Time period of rotation of earth about its axis $T = 24$ hr $= 86400$ seconds

Thus angular velocity of rotation of earth $w = \frac{2 π}{T} = \frac{2 π}{86400} = 7.3 \times 10^{- 5}$ rad/s

Instantaneous velocity of a point at the equator of earth's surface $v = r w = 7.3 \times 10^{- 5} \times 6.4 \times 10^{6} = 467.2$ m/s

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The radius of the Earth is approximately 6.4×106m. The instantaneous velocity of a point on the surface of the Earth at the equator is 4.672×10x. Find the value of x :

Radius of earth r=6.4×106 mTime period of rotation of earth about its axis T=24 hr =86400 secondsThus angular velocity of rotation of earth w=2πT=2π86400=7.3×10−5 rad/sInstantaneous velocity of a point at the equator of earth's surface v=rw=7.3×10−5×6.4×106=467.2 m/s

The radius of the Earth is approximately $6.4 \times 10^{6} m$ . The instantaneous velocity of a point on the surface of the Earth at the equator is $4.672 \times 10^{x}$ . Find the value of x :