<h2>
Answer:</h2>
Luke does a work of 308Nm
<h2>
Step-by-step explanation:</h2>
If we have a constant force 
that acts on a body in the same direction as the displacement 
, then the work 
is defined as the product of the force magnitude 
and the displacement magnitude 
. In other words:
, which is valid for a constant force in direction of straight-line displacement.
In this problem:
Therefore:
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Q in (-oo:+oo)
2/3 = (1/3)*q // - (1/3)*q
2/3-((1/3)*q) = 0
ddddddddd
d d
d d
(-1/3)*q+2/3 = 0 d d
d d
2/3-1/3*q = 0 // - 2/3 d d
d d
-1/3*q = -2/3 // : -1/3 d d
d d
q = -2/3/(-1/3) ddddddd dddddddd
dd dd
q = 2 dd dd
dd dddd dd
q = 2 dddddddddd dddddddddddd
Answer:
Step-by-step explanation:
- First, lets get rid of the fraction. I did this by multiplying both sides by
.
- We want to isolate
on one side of this equation. Let's put any values with on one side of the equation, and normal integers on the other.
- Divide both sides by
.
- If your teacher wants you to leave your final answer as an improper fraction, your final answer is this. If they want it to be a mixed number or decimal, your final answer will be:
