If the rocket it heavy and tall the distance wouldn't go very far but if the rocket was little and had no weight on it, it would go farther than the heavier one because of density/mass in the rocket
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A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).
Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.
Answer:
the work done is -88 J
Explanation:
Given the data in the question;
we know that;
Work done = F × S
where constant force F = ( 6i + 8j - 6k )
S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )
S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )
S = ( -12I + 7j + 12k )
so
Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )
Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )
Work force = -72 + 56 - 72
Work force = -88 J
Therefore, the work done is -88 J
The resulting net force is 650N
Explanation:
The force caused by air resistance against the plane's motion is called the Drag force. It acts against the motion of the plane. However, tailwinds push the plane from its tail and they apply a force is the direction of the moving plane.
The resulting force is the sum of the two forces; 450N +200N = 650N
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Drag and tail force on a plane :brainly.com/question/11922080
Keywords : thrust, force, tailwind, resulting net force
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Answer:
1e , 2j , 3c , 4d , 5k , 6a , 7i , 8b , 9m , 10h ,11g , 12f , 13l .
