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evablogger [386]
2 years ago
6

What's the best way to find the load capacity of a crane?

Engineering
1 answer:
Korvikt [17]2 years ago
5 0

Answer:

weight

Explanation:

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A bar having a length of 5 in. and cross-sectional area of 0. 7 in.2 is subjected to an axial force of 8000 lb. If the bar stret
andrew11 [14]

The modulus of elasticity is 28.6 X 10³ ksi

<u>Explanation:</u>

Given -

Length, l = 5in

Force, P = 8000lb

Area, A = 0.7in²

δ = 0.002in

Modulus of elasticity, E = ?

We know,

Modulus of elasticity, E = σ / ε

Where,

σ is normal stress

ε is normal strain

Normal stress can be calculated as:

σ = P/A

Where,

P is the force applied

A is the area of cross-section

By plugging in the values, we get

σ = \frac{8000 X 10^-^3}{0.7}

σ = 11.43ksi

To calculate the normal strain we use the formula,

ε = δ / L

By plugging in the values we get,

ε = \frac{0.002}{5}

ε = 0.0004 in/in

Therefore, modulus of elasticity would be:

E = \frac{11.43}{0.004} \\\\E = 28.6 X 10^3 ksi

Thus, modulus of elasticity is 28.6 X 10³ ksi

6 0
2 years ago
The chart shows the bids provided by four engineers to test a prototype.
klasskru [66]

Answer:

D

Explanation:

To know which is most or least cost-effective, it's not enough to look at only the per day rate, or only the time to complete.  You have to multiply them to get the total cost of the project.

\left[\begin{array}{ccccc}&Cost\ per\ day\ (\$)&Time\ to\ complete\ (days)&Total\ cost\ (\$)\\Zoe&500&8&4000\\Greg&650&10&6500\\Orion&400&12&4800\\Jin&700&5&3500\end{array}\right]

As you can see, Greg is the least cost-effective because he charges the most for the project.

8 0
3 years ago
YO CUTIE! HELP MEEEEEE! Hazardous waste is a category that includes all of the following EXCEPT
madam [21]

your answer should be <u>B. Soluble</u>

3 0
2 years ago
Read 2 more answers
Read two numbers from user input. Then, print the sum of those numbers. Hint -- Copy/paste the following code, then just type co
Softa [21]

Answer:

I am Providing Answer in C Language Program.

Explanation:

Please find attachment regarding code of taking two numbers input and adding them.

I would like to recommend you please use software which supports C language.

#include <stdio.h>

int main () {

int a, b, sum;

printf ("\ nEnter two no:");

scanf ("% d% d", & d, & e);

sum1 = d + e;

printf ("Sum:% d", sum1);

return (0);

}

4 0
3 years ago
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

8 0
3 years ago
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