The effiency of a machine is
(output work or energy) / (input work or energy) .
For the system described in the question, that's
(123 J) / (150 J) = 0.82 = 82% .
Positive electric charge and found inside the nucleus
Answer:
5.92×10⁷ J
Explanation:
We'll begin by converting 15 tons to Newton. This can be obtained as follow:
1 ton = 9806.65 N
Therefore,
15 ton = 15 ton × 9806.65 N / 1 ton
15 ton = 147099.75 N
Next, we shall convert one-quarter (¼) or 0.25 mile to metre. This can be obtained as follow:
100 mi = 160934 m
0.25 mi = 0.25 mi × 160934 / 100 mi
0.25 mi = 402.335 m
Finally, we shall determine the Workdone. This can be obtained as follow:
Force (F) = 147099.75 N
Distance (d) = 402.335 m
Workdone (Wd) =?
Wd = F × d
Wd = 147099.75 × 402.335
Wd = 5.92×10⁷ J
Thus, the Workdone is 5.92×10⁷ J
This is because of of the heating effect of a current. The glow is a result of current passing through the filament. The current experiences resistance as a result heat is generated. When resistance is at zero, there potential differences is not needed hence temperature generated will be at a constant.
Ok? I don’t know what you want me to do though
