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3 years ago

E = 50V, R= 1000 ohms, then I =

Engineering

1 answer:

Margarita [4]3 years ago

8 0

Explanation:

Current (I)= Voltage (V)

 Resistance (Ω). 

I= 50V

 1000(Ω)

I= 5V

 100(Ω)

I= 0.05 (Amp). Answer

Voltage=V

Current=I

Resistance=(Ω). (Ω)=Ohm

And SI unit of current is. Ampere(Amp) 

hope this helps you

have a good day.

 

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What are the characteristics of Polyamide?

Tems11 [23]

The two most important polyamides are poly(hexamethylene adipamide) (Nylon 6,6) and polycaprolactam (Nylon 6). Both have excellent mechanical properties including high tensile strength, high flexibility, good resilience, low creep and high impact strength (toughness).

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3 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

kirill [66]

Answer:

105.70 mm

Explanation:

Poisson’s ratio, v is the ratio of lateral strain to axial strain.

E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus

Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio

$\begin{array}{l}\\65.5{\rm{ GPa}} = 2\left( {25.4{\rm{ GPa}}} \right)(1 + \upsilon )\\\\\upsilon = 0.2893\\\end{array}$

Original length $L_(i}$

$\upsilon = - \left( {\frac{{\left( {\frac{{{d_f} - {d_i}}}{{{d_i}}}} \right)}}{{\left( {\frac{{{L_f} - {L_i}}}{{{L_i}}}} \right)}}} \right)$

Where $d_{f}$ is final diameter, $d_{i}$ is original diameter, $L_{f}$ is final length and $L_{i}$ is original length.

$\begin{array}{l}\\0.2893 = - \left( {\frac{{\left( {\frac{{30.04{\rm{ mm}} - {\rm{30 mm}}}}{{{\rm{30 mm}}}}} \right)}}{{\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right)}}} \right)\\\\\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right) = - 4.6088 \times {10^{ - 3}}\\\end{array}$

$\begin{array}{l}\\105.2 - {L_i} = - \left( {4.6088 \times {{10}^{ - 3}}} \right){L_i}\\\\105.2 = 0.9953{L_i}\\\\{L_i} = 105.70{\rm{ mm}}\\\end{array}$

Therefore, the original length is 105.70 mm

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4 years ago

Consider the following incomplete code segment, which is intended to print the sum of the digits in num. For example, when num i

ValentinkaMS [17]

Answer:

A) while (num >= 0)

Explanation:

To understand why we need to focus on the module and division operation inside the loop. num % 10 divide the number by ten and take its remainder to then add this remainder to sum, the important here is that we are adding up the number in reverse order and wee need to repeat this process until we get the first number (1%10 = 1), therefore, num need to be one to compute the last operation.

A) this is the correct option because num = 1 > 0 and the last operation will be performed, and after the last operation, num = 1 will be divided by 10 resulting in 0 and 0 is not greater than 0, therefore, the cycle end and the result will be printed.

B) This can not be the option because this way the program will never ends -> 0%10 = 0 and num = 0/10 = 0

C) This can not be the option because num = 1 > 1 will produce an early end of the loop printing an incomplete result

D) The same problem than C

E) There is a point, before the operations finish, where sum > num, this will produce an early end of the loop, printing an incomplete result

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3 years ago

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erastovalidia [21]

So 20 plus 60 = 40 then

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3 years ago

The way most recursive functions are written, they seem to be circular at first glance, defining the solution of a problem in te

EastWind [94]

Question Continuation

int factorial(int n) {

if(n == 0)

return 1;

else

return n * factorial(n - 1);

}

Provide a brief explanation why this recursive function works.

Show all steps involved in calculating factorial(3) using the function deﬁned.

Answer:

1. Brief explanation why this recursive function works.

First, the recursive method factorial is defined.

This is the means through with the machine identifies the method.

The method is defined as integer, the machine will regard it as integer.

When the factorial is called from anywhere that has access to it, which in this case is within the factorial class itself. This means you can call it from the main method, or you can call it from the factorial method itself. It's just a function call that, well, happens to call itself.

2. Steps to calculate factorial(3)

1 First, 3 is assigned to n.

2. At line 2, the machine checks if n equals 0

3. If yes, the machine prints 1

4. Else; it does the following from bottom to top

factorial(3):

return 3*factorial(2);

return 2*factorial(1):

return 1;

Which gives 3 * 2 * 1 = 6

5. Then it prints 6, which is the result of 3!

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3 years ago