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erik [133]
3 years ago
14

1. A thin-walled cylindrical pressure vessel is capped at the end and is subjected to an internal pressure (p). The inside diame

ter of the vessel is 6 ft and the wall thickness is 1.5 inch. The vessel is made of steel with tensile yield strength and compressive yield strength of 36 ksi. Determine the internal pressure required to initiate yielding according to (a) The maximum-shear-stress theory of failure, and (b) The maximum-distortion-energy theory of failure, if a factor of safety (FS) of 1.5 is desired.
Engineering
1 answer:
Vesna [10]3 years ago
6 0
I DONT KNOW OKAY UGHHH
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Can someone answer plz!! It’s 24 points
fgiga [73]

Explanation:

750 microvolt is your answer

please mark as brilliant

3 0
3 years ago
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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
4 years ago
Problem the pressure at a given point is 50 mmhg absolute
VLD [36.1K]

Answer:

6.65 kPa.

- 73.3 kPa.

Explanation:

Without much ado let's jump right into the solution to the problem given. It is given that the pressure = 50 mmHg and the solution to this question is to write out the value of the pressure in kpa and kPa gauge.

P(a) = 0.05m × 133 kN/m³ = 6.65 kPa.

The P(gauge) =( [ 0.05 × 13.6 × 9810] ÷ 1000 ) - 80 = - 73.3 kPa.

7 0
3 years ago
When conducting an immediate charge on 2023 ariya, what is the first step if the timer is set?.
avanturin [10]

The first step to perform when conducting an immediate charge is to press the immediate charge switch on the lower left dash.

<h3>What is an immediate charge?</h3>

This is a vehicle component that allows a quick control of the machine.

However, the first step to achieve when executing an immediate charge is to press the immediate charge switch on the more low left dash.

Read more about immediate charge

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Can space debris take out a whole state
vodomira [7]
Yes depends how big the debris is
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