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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
Answer:
Explanation:
It is given that,
Mass of bundle of shingles, m = 10 kg
Upward acceleration of the shingles, 
The radius of the motor of the pulley, r = 0.17 m
Let T is the tension acting on the shingles when it is lifted up. It can be calculated as :
T = 113 N
Let 
is the minimum torque that the motor must be able to provide. It is given by :
So, the minimum value of torque is 19.21 N-m. Hence, this is the required solution.
Answer:
go to the link quizzlet it will give you tha answer
Explanation:
Ahhh, there aren’t option to work with :v
Try writing them down here thnx
