Calculate the magnitude of the linear momentum \( p_0 \) of a 33.6 kg object on the Equator, as measured from the center of Earth. The radius \( R \) of Earth is 6371 km.\[ p_0 = \] \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ kg·m/sDefine the time at which the previous result was calculated as \( t_0 \). What is the magnitude of the net change \( \Delta p_1 \) in the object's momentum in the 6 h following \( t_0 \)?\[ \Delta p_1 = \] \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ kg·m/sNow, calculate the magnitude of the total change \( \Delta p_2 \) in the object's momentum 12 h after \( t_0 \).\[ \Delta p_2 = \] \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ kg·m/s

First we will calculate the speed of rotation of the earth using the radius of earth and the time…

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