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2 years ago

7

engineering uses data from pareto charts to analyse motor faults caused during their production. Explain one advantage of using

pareto charts in this situation. (4 Marks) Please help

1 answer:

lions [1.4K]2 years ago

3 0

The advantage of a pareto chart is to make sure they have all of their tools

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The goal of the following model is generate a clock waveform that has the clock high 4 time units and low 4 time units, with the

DedPeter [7]

Answer:

module q2;

reg clock;

initial begin

clock = 0;

#20 clock = 1; // at 20 set high

always begin

#4 clock = ~clock; // then every four units change

end

end

endmodule

3 0

2 years ago

Rolling and Shearing are the types of a)-Bulk Deformation Process b)- Sheet Metal Process c)- Machining Process d)- Both a &

Margarita [4]

Answer:

a)Bulk deformation process

Explanation:

<u>Rolling</u>

Rolling is a metal forming process.In rolling work piece passes through two moving rollers and get compressed.in rolling thickness of work piece will reduces and length of work piece will increase for maintaining the constant area.Due to compression bulk deformation takes place.

<u>Shearing</u>

In shearing one surface slides on another surface and deformation take place.shearing is a machining process.This is also a bilk motion deformation process.

So from above we can say that option a is right.

5 0

2 years ago

Mining is an example of this type of business

luda_lava [24]

Answer:

Mining would go under Industry organization.

5 0

2 years ago

An aluminium alloy bar of diameter 12.5 mm and length 27 m loaded in uniaxial tension to a force of 3 kN. Determine the length o

KonstantinChe [14]

Answer:

27009.56 mm

Explanation:

Given:

Diameter of the aluminium alloy bar, d = 12.5 mm

Length of the bar, L = 27 m = 27 × 10³ mm

Tensile force, P = 3 KN = 3 × 10³ N

Elastic modulus of the bar, E = 69 GPa = 69 × 10³ N/mm²

Now,

for the uniaxial loading, the elongation or the change in length (δ) due to the applied load is given as:

$\delta=\frac{PL}{AE}$

where, A is the area of the cross-section

$A=\frac{\pi d^2}{4}$

or

$A=\frac{\pi\times12.5^2}{4}$

or

A = 122.718 mm²

on substituting the respective values in the formula, we get

$\delta=\frac{3\times10^3\times27\times10^3}{122.718\times69\times10^3}$

or

δ = 9.56 mm

Hence, the length after the force is applied = L + δ = 27000 + 9.56

= 27009.56 mm

7 0

2 years ago

What is the process pf distributing and selling clean fuel?

kipiarov [429]

Answer:

I got hygition production and decomposing

Explanation:

6 0

2 years ago