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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?

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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Two piezometers have been placed along the direction of flow in a confined aquifer that is 30.0 m thick. The piezometers are 280

Dafna11 [192]

Answer:

time = 224 days

Explanation:

given data

thick = 30 m

piezometers = 280 m

head between two = 1.4 m

aquifer hydraulic conductivity = 50 m

porosity = 20%

solution

we get here average pole velocity that is get by using Darcy law that is

Va = $\frac{k}{\eta } \times \frac{\Delta h}{L}$ ................1

here Va is average pole velocity and k is hydraulic conductivity and $\eta$ is porosity

here v is = $k \times \frac{dh}{dl}$ ...........2

v = 50 × $\frac{1.4}{280}$

v = 0.25 m/day

and here average linear velocity Va will be

Va = $\frac{v}{\eta }$

Va = $\frac{0.25}{0.2}$

Va = 1.25 m/day

travel time for water will be

time = $\frac{280}{1.25}$

time = 224 days

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2 years ago

Various factors to be considered in deciding the factor of safety?

Eduardwww [97]

Answer with Explanation:

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1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.

2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.

3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.

4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.

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2 years ago

A. For a 200g load acting vertically downwards at point B’, determine the axial load in members A’B’, B’C’, B’D’, C’D’ and C’E’.

Nonamiya [84]

Answer:

attached below

Explanation:

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2 years ago

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing

Viktor [21]

Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that $\mu = 0.8$

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

$P(X \leq 4) + P(X > 4) = 1$

We want P(X > 4). Then

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

$P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595$

$P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438$

$P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383$

$P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014$

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2 years ago

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AysviL [449]

Answer is your company’s address

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