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Bumek [7]
3 years ago
8

Corrections for curvature and refraction (c r) are applied to: ____________.

Engineering
1 answer:
Y_Kistochka [10]3 years ago
8 0

Answer:

b.

Explanation:

Foresight rod reading is any measurement taken at a given sight to determine a particular elevation. The backsight rod readings are usually used to a point of certain elevation. It is added to the elevation to determine the height of the instrument. These are phenomena are used in differential leveling and are applied to corrections for curvature and refraction.

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Assume a function requires 20 lines of machine code and will be called 10 times in the main program. You can choose to implement
Volgvan

Answer:

"Macro Instruction"

Explanation:

A macro definition is a rule or pattern that specifies how a certain input sequence should be mapped to a replacement output sequence according to a defined procedure. The mapping process that instantiates a macro use into a specific sequence is known as macro expansion.

It is a series of commands and actions that can be stored and run whenever you need to perform the task. You can record or build a macro and then run it to automatically repeat that series of steps or actions.

7 0
3 years ago
g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
3 years ago
I need ideas for what to build because I have some spare wood.
Misha Larkins [42]

Answer:

small guitar with no strings?

Explanation:

it would be fun to make i think

6 0
3 years ago
The best grade of hardwood lumber that is generally available is _____​
Vesnalui [34]

Answer:

FAS

Explanation:

first and second grade

5 0
2 years ago
A 3 m aluminum pole is kept at a residential site for construction
Aliun [14]

Answer:

I don't know sorry

Explanation:

5 0
3 years ago
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