Answer:
The required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
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Explanation:
We are given current density of 459 A/cm² and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.
Recall that current density is given by
j = I/A
where I is the current flowing through the wire and A is the area of the wire
A = πr²
but r = d/2 so
A = π(d/2)²
A = πd²/4
so the equation of current density becomes
j = I/πd²/4
j = 4I/πd²
Re-arrange the equation for d
d² = 4I/jπ
d = √4I/jπ
d = √(4*0.53)/(459π)
d = 0.0383 cm
Therefore, the required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
.
Wait what does Jessica want ?
Answer:
429.5 ft
Explanation:
We are told that;
1 ft = 0.305 meters
We want to convert 131 meters to ft, by proportion we have;
Distance = (131 × 1)/0.305
Distance = 429.5082 ft
We want to approximate to the nearest tenth of a foot.
This gives us;
Distance = 429.5 ft
