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horrorfan [7]
3 years ago
6

What is the output of a system having the transfer function G = 2/[(s + 3) x(s + 4)] and subject to a unit impulse?

Engineering
1 answer:
lisov135 [29]3 years ago
8 0

Answer:

output=\frac{2}{(s+3)(s+4)}

Explanation:

output =transfer function ×input

here transfer function G=\frac{2}{(S+3)(S+4)} {}

input = unit impulse

in S domain unit impulse =1

so output =\frac{2}{(S+3)(S+4)} {}

=\frac{2}{(s+3)(s+4)}

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The yield strength of mild steel is 150 MPa for an average grain diameter of 0.038 mm ; yield strength is 250 MPa for average gr
djyliett [7]

Answer:

Explanation:

Hall-Petch equation provides direct relations between the strength of the material and the grain size:

σ=σ0+k/√d , where d- grain size, σ- strength for the given gran size, σ0 and k are the equation constants.

As in this problem, we don't know the constants of the equation, but we know two properties of the material, we are able to find them from the system of equations:

σ1=σ0+k/√d1

σ2=σ0+k/√d2 , where 1 and 2 represent 150MPa and 250MPa strength of the steel.

Note, that for the given problem, there is no need to convert units to SI, as constants can have any units, which are convenient for us.

From the system of equations calculations, we can find constant: σ0=55.196 MPa, k=18.48 MPa*mm^(0.5)

Now we are able to calculate strength for the grain diameter of 0.004 mm:

σ=55.196+18.48/(√0.004)=347.39 MPa

The strength of the steel with the grais size of 0.004 mm is 347.39 MPa.

6 0
4 years ago
Write definitions for the following two functions:______
lawyer [7]

Answer:

<em>Written in Python</em>

def SumN(n):

     total = 0

     for i in range(1,n+1):

           total = total + i

     print("Total: ",total)

def SumNCubes(n):

     total = 0

     for i in range(1,n+1):

           total = total + i**3

     print("Cube: ",total)

n = int(input("User Input: "))

if n > 0:

     SumN(n)

     SumNCubes(n)

Explanation:

The SumN function is defined here

def SumN(n):

This line initializes the total to 0

     total = 0

The following iteration compute the required sum

<em>      for i in range(1,n+1): </em>

<em>            total = total + i </em>

This line outputs the calculated sum

     print("Total: ",total)

The SumNCubes function is defined here

def SumNCubes(n):

This line initializes the total to 0

     total = 0

The following iteration compute the required sum of cubes

<em>      for i in range(1,n+1): </em>

<em>            total = total + i**3 </em>

This line outputs the calculated sum of cubes

     print("Cube: ",total)

The main starts here; The first line prompts user for input

n = int(input("User Input: "))

The next line checks if input is greater than 0; If yes, the two defined functions are called

if n > 0:

     SumN(n)

     SumNCubes(n)

6 0
3 years ago
A slug travels 3 centimeters in 3 seconds. A snail travels 6 centimeters in 6 seconds. Both travel at constant speeds. Mai says,
Irina-Kira [14]

Answer:

i dont agree with mai because they were both going 1cm per second

Explanation:

3÷3=1

6÷6=1

they both are difrent numbers but equal the same thing

8 0
4 years ago
An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100
Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}

Where:

\sigma_{m} - Mean stress, measured in megapascals.

\sigma_{min} - Minimum stress, measured in megapascals.

\sigma_{max} - Maximum stress, measured in megapascals.

If we know that \sigma_{min} = -100\,MPa and \sigma_{max} = 300\,MPa, the mean stress is:

\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}

\sigma_{m} = 100\,MPa

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

\sigma_{a} = |\sigma_{max}-\sigma_{min}|

Where \sigma_{a} is the stress amplitude, measured in megapascals.

If we know that \sigma_{min} = -100\,MPa and \sigma_{max} = 300\,MPa, the stress amplitude is:

\sigma_{a} = |300\,MPa-(-100\,MPa)|

\sigma_{a} = 400\,MPa

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio (R) is the ratio of the stress amplitude to mean stress. That is:

R = \frac{\sigma_{a}}{\sigma_{m}}

If we know that \sigma_{m} = 100\,MPa and \sigma_{a} = 400\,MPa, the stress ratio is:

R = \frac{400\,MPa}{100\,MPa}

R = 4

The stress ratio of the aluminium part is 4.

3 0
3 years ago
a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8
djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam  along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we  determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = \frac{MC}{I}  =  \frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam =  159.07 / 2 = 79.535 MPa  

attached below is the remaining solution

6 0
3 years ago
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