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2 years ago

Component of earthing and reasons why each material is being used

Engineering

1 answer:

timofeeve [1]2 years ago

3 0

Answer:

The components of earthing are Earth electrode, Main earthing terminals/bars, Earthing conductors, Protective conductors Equipontential binding conductors and Electrically independent electrodes.

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Please define the coefficient of thermal expansion?

Vikki [24]

Answer:

The coefficient of thermal expansion tells us how much a material can expand due to heat.

Explanation:

Thermal expansion occurs when a material is subjected to heat and changes it's shape, area and volume as a result of that heat. How much that material changes is dependent on it's coefficient of thermal expansion.

Different materials have different coefficients of thermal expansion (i.e. It is a material property and differs from one material to the next). It is important to understand how materials behave when heated, especially for engineering applications when a change in dimension might pose a problem or risk (eg. building large structures).

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3 years ago

If the price of the car is less than or equal to your available cash, display "no". If the price of the car is more than your av

Ede4ka [16]

Answer:

function decision(car_price, available_cash) {

if(car_price <= available_cash) {

console.log("no");

}

else {

console.log("yes");

}

decision(car_price, available_cash); or decision(available_cash, car_price);

Explanation:

using functions in Javascript:

functions; this refers to dividing codes into reusable parts.

e.g function function_name() {

console.log("How are you?");

}

you can call or invoke this function by using its name followed by parenthesis, like this: function_name(). each time the function is called it will print out "How are you?".

Parameters: these are variables that act as placeholders for the values that are to be input into a function when it is called

Arguments: The actual values that input or passed into a function when it is called.

e.g

function function_name(parameter1, parameter2) {

console.log(parameter1, parameter2);

}

then we call function_name: function_name("please", "leave"):we have passed two arguments, "please" and "leave". Inside the function parameter1 equals "please" while parameter2 equals "leave".

Hence, from the question given the two parameters "car_price" and "available_cash" respectively, we write the function with name function_name:

function decision(car_price, available_cash) {

if(car_price <= available_cash) {

console.log("no");

}

else {

console.log("yes");

}

decision(car_price, available_cash); or decision(available_cash, car_price);

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3 years ago

Please help me do this with my exam tomorrow.

blagie [28]

Answer:

ddddddddddddddddddddddddddddd

Explanation:

cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

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2 years ago

18) Technician A says that adjustable wrenches should be used when

kykrilka [37]

Tryna boost my score for college stuff could you give me the brainiest and a thanks? Hope you find your answer your looking for!

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2 years ago

I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have

drek231 [11]

Answer:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = $\lambda e^{-\lambda t}$

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

$P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1$

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

$p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [ -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905$

$P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!} = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!} = 3.915\times 10^{-12}$

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) $CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}$ .

4 0

3 years ago

Component of earthing and reasons why each material is being used​

Component of earthing and reasons why each material is being used