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1 year ago

Exercise 6.4.8: Sum Two Number

Engineering

1 answer:

kaheart [24]1 year ago

3 0

Below is the program with function that takes two arguments and returns their sum.

Coding part:

def add_num(a1,b1): /*function for addition*/

sum=a1+b1

return sum /*return value*/

num1z1=33 /*variable declaration*/

num2z2=78

print("The sum is",add_num(num1z1,num2z2)) /*call the function*/

What is Function?

A function is a piece of code that performs a specific task. It is callable and reusable multiple times. You can pass data to a function, and it can return data to you. Many programming languages include built-in functions that can be found in their libraries, but you can also write your own.

To learn more about Function, visit: brainly.com/question/17216645

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An incoming signal is at a frequency of 500kHz. This signal needs to be acquired and all other signals attenuated. Design a pass

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Answer:

C_h = 0.166 nF

C_L = 0.153 nF

Explanation:

Given:

- Ideal frequency f_o = 500 KHz

- Bandwidth of frequency BW = 40 KHz

- The resistance identical to both low and high pass filter = 2 Kohms

Find:

Design a passive band-pass filter to do this by cascading a low and high pass filter.

Solution:

- First determine the cut-off frequencies f_c for each filter:

f_c,L for High pass filter:

f_c,L = f_o - BW/2 = 500 - 40/2

f_c,L = 480 KHz

f_c,h for Low pass filter:

f_c,h = f_o + BW/2 = 500 + 40/2

f_c,h = 520 KHz

- Now use the design formula for R-C circuit for each filter:

General design formula:

f_c = 1 /2*pi*R*C_i

C,h for High pass filter:

C_h = 1 /2*pi*R*f_c,L

C_h = 1 /2*pi*2000*480,000

C_h = 0.166 nF

C,L for Low pass filter:

C_L = 1 /2*pi*R*f_c,h

C_L = 1 /2*pi*2000*520,000

C_L = 0.153 nF

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3 years ago

Consider a unidirectional continuous fiber-reinforced composite with epoxy as the matrix with 55% by volume fiber.i. Calculate t

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Answer:

I)E= 40.95 GPa

II)E=5.29 GPa

Explanation:

Given that

E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45

E₂ = 72.5 GPa ,V₂=0.55

Longitudinal moduli given as ;

E= E₁V₁+E₂V₂

E= 2.41 x 0.45 + 72.5 x 0.55 GPa

E= 40.95 GPa

II)

E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45

E₂ =230 GPa ,V₂=0.55

Transverse moduli given as:

$\dfrac{1}{E}=\dfrac{V_1}{E_1}+\dfrac{V_2}{E_2}$

$\dfrac{1}{E}=\dfrac{0.45}{2.41}+\dfrac{0.55}{230}$

E=5.29 GPa

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Answer:

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The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

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