0N. The net force acting on this firework is 0.
The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.
Based on the free-body diagram, we have:
The force of gases is Fgases = 9,452N
The force of the rocket Frocket = -9452
Then, the net force acting is:
Fnet = Fgases + Frocket
Fnet = 9,452N - 9,452N = 0N
I'm not quite sure what happens to Fay so I didn't finish but hope it helps
The correct answer is the reverse wave I took the test
5m
Explanation:
Given parameters:
Weight of object = 50N
Work done in lifting object = 250J
Unknown:
Vertical height = ?
Solution:
The work done on an object is the force applied to lift a body in a specific direction.
Work done = force x distance
Weight is a force in the presence of gravity;
Work done = weight x height of lifting
Height of lifting = 
Height of lifting =
= 5m
The vertical height through which the object was lifted is 5m
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