Answer: See below
Explanation:
Part (a):
The angular speed of the Earth can be calculated as,
[tex]\omega &= \dfrac{{2\pi }}{T}\\ \omega &= \dfrac{{2\pi }}{{\left( {23 \times 3600} \right)\left( {56 \times 60} \right) + 4}}\\ \omega &= 7.29 \times {10^{ - 5}}\;{\rm{rad/s}}[/tex]
Part (b):
The linear velocity of the object can be calculated as,
[tex]v &= \omega r\sin \theta \\ v &= \left( {7.29 \times {{10}^{ - 5}}} \right)\left( {6.37 \times {{10}^6}} \right)\sin 47^\circ \\ v &= 339.62\;{\rm{m/s}}[/tex]
Part (c):
The acceleration of the object can be calculated as,
[tex]a &= \dfrac{{{v^2}}}{{r\sin \theta }}\\ a &= \dfrac{{{{339.62}^2}}}{{6.37 \times {{10}^6}\sin 47^\circ }}\\ a &= 0.024\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}[/tex]
