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Nuetrik [128]
3 years ago
7

What are the dimension of the container​

Engineering
1 answer:
zmey [24]3 years ago
4 0

Answer:

Dimensions of a standard container is length 5.440 m 17'10 3/16' width 2.294 m 7'6 1/4' height centimeters 3.237 7'4 1/16'

Explanation:

Standard height for a shipping container is 8 feet 6 inches

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Question 7 of 10
exis [7]

Answer: Define the problem

Explanation: I just took the test

8 0
4 years ago
Print "userNum1 is negative." if userNum1 is less than 0. End with newline. Assign userNum2 with 2 if userNum2 is greater than 1
miv72 [106K]

Answer:

From the question, we have two variables

1. userNum1

2. userNum2

And we are to print "userNum1 is negative" if userNum1 is less than 0.

Then Assign userNum2 with 2 if userNum2 is greater than 10.

Otherwise, print "userNum2 is less or equal 10.".

// Program is written in C++ Programming Language

// Comments are used for explanatory purpose

// Program starts here

#include<iostream>

using namespace std;

int main ()

{

// Declare variables

int userNum1, userNum2;

// Accept input for these variables

cin>>userNum1, userNum2;

// Condition 1

if(userNum1 < 0)

{

cout<<"userNum1 is negative"<<'\n';

}

// Condition 2

if(userNum2 > 10)

{

userNum2 = 2;

}

// If condition is less than 10

else

{

cout<<"userNum2 is less or equal to 10"<<\n;

}

return 0;

}

// End of Program.

3 0
3 years ago
Drag each item to show if it is an element or not an element.
Flauer [41]

Answer:

CARBON

Explanation:

HOPE THIS HELPS SORRY FOR CAPS

5 0
3 years ago
The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr
Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

(\sigma_ x)  = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}

Since d = 2r

Then:

(\sigma_ x)  = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}

(\sigma_ x)  = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}

(\sigma_ x)  = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}

(\sigma_ x)  = 11917000.38

(\sigma_ x)  = 11.917 \times 10^6 \ Pa

(\sigma_ x)  = 11.917 \ MPa

\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\  \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

\tau _{xy} =0

3 0
3 years ago
What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C
Anvisha [2.4K]

Answer:

331809.5gallon/hr or 92.16gallon/s

Explanation:

What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C

convert 9.4 acre to inches we have=5.896*10^7

How to calculate Peak runoff discharge

1. take the dimension of the roof

2. multiply the dimension by the n umber of inches of rainfall

3. Divide by 231 to get gallon equivalence (because 1 gallon = 231 cubic inches)

5.896*10^7*1.3

7.66*10^7 cubic inches/hr

1 gallon=231 cubic inches

7.66*10^7 cubic inches=331809.5gallon/hr or 92.16gallon/s

this is gotten by converting 1 hr to seconds

331809.5gallon/hr /3600s=92.16gallon/s

8 0
3 years ago
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