A spacecraft descends vertically near the surface of Planet X. An upward thrust

Question

A spacecraft descends vertically near the surface of Planet X. An upward thrust of 25.0 from its engines slows it down at a rate of 1.20, but if an upward thrust of only 10.0 is applied, it speeds up at a rate of 0.80. What is the direction of the acceleration of the spacecraft in the case of 25.0 thrust? Answer is Upward. Draw a free-body diagram for the spacecraft in the case of 25.0 thrust? Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. What is the direction of the acceleration of the spacecraft in the case of 10.0 thrust? Answer is Downward. Draw a free-body diagram for the spacecraft in the case of 10.0 thrust? Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. I need the answers to the diagrams. Please

Explanation / Answer

F=m*a and m is constant on any planet 25000-m*g=m*1.2 10000-m*g=-m*0.80 m*g is the weight 25000/1.2-m*g/1.2=m 10000/0.80-m*g/0.80=-25000/1.2+m*g/1.2 solve for m*g m*g=(10000/0.80+25000/1.2)/ (1/1.2+1/0.80) 16 kN

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