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Nostrana [21]
2 years ago
6

Suppose that the material to be used in a fuse melts when the current density rises to 459 A/cm2. What diameter of cylindrical w

ire should be used to limit the current to 0.53 A?
Engineering
2 answers:
slega [8]2 years ago
7 0

Answer:

diameter is 1 cm

Explanation:

diameter is 1 cm

I am Lyosha [343]2 years ago
6 0

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 .

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 .

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3 years ago
The tolerance for a geometric can be easily identified by reading the _________.
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Explanation:

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alexgriva [62]

Answer:

metals, composite, ceramics and polymers.

Explanation:

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ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

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Answer:

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Explanation:

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Diameter of pipe is 10mm.

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Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is 1.557\times10^{-7}m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

Re=\frac{vd}{\nu}

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