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-Dominant- [34]
2 years ago
5

Design a program that calculates the area and circumstance of rectangle?​

Engineering
1 answer:
Phantasy [73]2 years ago
4 0
Let “w” and “L” be the width and length of the rectangle. “p” and “a” are perimeter and area
For python,
w=int(input(“width”))
l=int(input(“length”))
a= w*l
p=2*w+2*l
print(str(a), str(p)
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Given the inherent costs of regulation it is safe to say that there is always a negative economic impact associated with regulat
Alexus [3.1K]

the answer is true.                                     <u>                    </u>                                                              

6 0
2 years ago
A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude i
Gre4nikov [31]

Answer:

13.4 mm

Explanation:

Given data :

Load amplitude ( F )  = 22,000 N

factor of safety ( N )= 2.0

Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa

<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>

minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm

<em>attached below is a detailed solution</em>

3 0
3 years ago
Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t
densk [106]

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t)  be a real value signal for x ( jw ) = 0 , |w| > 200\pi

g(t) = x ( t ) sin ( 2000 \pi t )

x_{1} (t) = \frac{1}{2}  x(t)  sin ( 4000\pi t )

next we apply Fourier transform

attached below is the remaining part of the solution

6 0
3 years ago
The state of plane strain on an element is:
balu736 [363]

Answer:

a. ε₁=-0.000317

   ε₂=0.000017

θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2}  \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2}  \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6}  \pm 1.67 \times 10^{-4}

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5

θ= -13.28° and  76.72°

To determine the direction of ε₁ and ε₂

\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2}  + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta  + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2}  + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56)  + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain

\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}

=3.335 *10^-4

\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )

ε(avg) =150 *10^-6

orientation of γmax

tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}

θ = 31.71 or -58.29

To determine the direction of γmax

\gamma _{x'y' }=  - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta  + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }=  - \frac{-300*10^{-6} - \ 0}{2} sin(63.42)  + \frac{150*10^{-6}}{2}cos(63.42)

= 1.67 *10^-4

4 0
4 years ago
A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m&gt;s. At the same inst
expeople1 [14]

Answer:

s= 20.4 m  

Explanation:

First lets write down equations for each ball:  

s=so+vo*t+1/2a_c*t^2

for ball A:

s_a=30+5*t+1/2*9.81*t^2

for ball B:  

s_b=20*t-1/2*9.81*t^2

to find time deeded to pass we just put that

s_a = s_b  

30+5*t-4.91*t^2=20*t-4.9*t^2

t=2 s  

now we just have to put that time in any of those equations an get distance from the ground:  

s = 30 + 5*2 -1/2*9.81 *2^2  

s= 20.4 m  

6 0
3 years ago
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