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lozanna [386]
4 years ago
15

What does the following program segment do? Declare Count As Integer Declare Sum As Integer Set Sum = 0 For (Count = 1; Count &l

t; 50; Count++) Set Sum = Sum + Count End For
Engineering
1 answer:
Troyanec [42]4 years ago
6 0

1225

<u>Explanation:</u>

This segment helps initialize sum as 0. The for loop is used to increment with every execution and it is added to the sum. The loop runs 49 times and every time the count is added to the sum. In short it is the sum of first 49 natural numbers i.e 1+2+3+......+49.

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In a steady flow device, the properties of the system remains constant with time. a)True b) False
Leviafan [203]

Answer:

True

Explanation:

By definition of steady flow we have

\frac{\partial f(x,y,z,t) }{\partial t}=0

where f(x,y,z,t) is any property of the system under consideration

=> f(x,y,z,t) = constant

7 0
3 years ago
Using the Tsai-Hill failure criterion, determine the strength of a lamina under equal biaxial tension and shear at 45o to the fi
Alinara [238K]

Answer:

Explanation:

solution to the question

5 0
3 years ago
Acoke can with inner diameter(di) of 75 mm, and wall thickness (t) of 0.1 mm, has internal pressure (pi) of 150 KPa and is suffe
NemiM [27]

Answer:

All 3 principal stress

1. 56.301mpa

2. 28.07mpa

3. 0mpa

Maximum shear stress = 14.116mpa

Explanation:

di = 75 = 0.075

wall thickness = 0.1 = 0.0001

internal pressure pi = 150 kpa = 150 x 10³

torque t = 100 Nm

finding all values

∂1 = 150x10³x0.075/2x0,0001

= 0.5625 = 56.25mpa

∂2 = 150x10³x75/4x0.1

= 28.12mpa

T = 16x100/(πx75x10³)²

∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]

= 1/2[84.37±√791.2969+5.827396]

= 1/2[84.37±28.33]

∂1 = 1/2[84.37+28.33]

= 56.301mpa

∂2 = 1/2[84.37-28.33]

= 28.07mpa

This is a 2 d diagram donut is analyzed in 2 direction.

So ∂3 = 0mpa

∂max = 56.301-28.07/2

= 14.116mpa

6 0
3 years ago
A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed i
kogti [31]

Answer

given,

Speed of vehicle = 65 mi/hr

                            = 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

   = 25050 ft

using super elevation formula

e + f = \dfrac{v^2}{rg}

0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}

r = \dfrac{95.33^2}{32.2\times 0.18}

  r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

L = \dfrac{\piR\Delta}{180}

L = \dfrac{\pi\times 1574\times 38}{180}

L = 1044 ft

Tangent of the curve calculation

  T = R tan(\dfrac{\Delta}{2})

  T = 1574 tan(\dfrac{38}{2})

      T = 542 ft

The station PC and PT are

 PC = PI - T

 PC = 25050 - 542

       = 24508 ft

       = 245 + 8 ft

PT = PC + L

     = 24508 + 1044

     =25552

     = 255 + 52 ft

the middle ordinate calculation

MO = R(1-cos\dfrac{\Delta}{2})

MO = 1574\times (1-cos\dfrac{38}{2})

     MO = 85.75 ft

degree of the curvature

D = \dfrac{5729.578}{R}

D = \dfrac{5729.578}{1574}

D = 3.64°

8 0
4 years ago
Two semiconductor materials have exactly the same properties except material A has a bandgap energy of 0.90 eV and material B ha
AlekseyPX

Answer: hello your question is incomplete attached below is the complete question

answer : Ac = 5° , A_{f} = 2.5°

Explanation:

Bandgap energy for material A = 0.90 eV

Bandgap energy for material B = 1.10 eV

<u>Calculate the ratio of ni for Material B to Material B </u>

Total derivation ( d ) = d1 + d2

 d = A_{c} ( μ_{c} - 1 ) + A_{f} ( μ_{f} - 1 )  ---- ( 1 )

where : d = 1° , μ_{c} = 1.5 , μ_{f} = 1.6

Input values into equation 1 above

1° = 0.5Ac + 0.6Af  ---- ( 2 )

also d = d1 [ 1 - w/ w1 ] ------ ( 3 )

∴ d = Ac ( μ_{c} - 1 ) ( 1 - w/w1 )

  1° = Ac ( 1.5 - 1 ) ( 1 - 0.06/0.1 ) --- ( 4 )

resolving  equation ( 4 )

Ac = 5°

resolving equation ( 2 )

A_{f} = 2.5°

8 0
3 years ago
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