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lana [24]
3 years ago
10

5/6+4(3/4g-6) in standard form

Mathematics
2 answers:
just olya [345]3 years ago
6 0

Answer:

3g- 139/6

Step-by-step explanation:

gavmur [86]3 years ago
3 0

Answer:3g- 139/6

Step-by-step 3g- 139/6

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The geometic mean of 4 and 21
-Dominant- [34]

Answer:

12.5

Step-by-step explanation:

4+21=25

25/2=12.5

4 0
3 years ago
A closed cylindrical vessel contains a fluid at a 5MPa pressure. The cylinder, which has an outside diameter of 2500mm and a wal
Julli [10]

Answer:

1) Increase in the diameter equals 3.5 mm

2) Increase in the length equals 0.0003724L_{i} where L_{i} is the initial length of the vessel.

Step-by-step explanation:

The diametric strain in the vessel is given by

\epsilon_{D} =\epsilon_{diam}-\nu \epsilon _{axial}

We have

\epsilon _{diam}=\frac{\sigma _{hoop}}{E}\\\\\sigma _{hoop}=\frac{\Delta P\times D}{2t}\\\\\therefore \epsilon _{diam}=\frac{\Delta P\times D}{2t\times E}

Applying values we get

\therefore \epsilon _{diam}=\frac{5\times 10^{6}\times 2.5}{2\times 20\times 10^{-3}\times 193\times 10^{9}}\\\\\therefore \epsilon _{diam}=\frac{5}{3088}

Similarly axial strain is given by

\epsilon _{diam}=\frac{\sigma _{axial}}{E}

\sigma _{axial}=\frac{\Delta P\times D}{4t}\\\\\therefore \epsilon _{axial}=\frac{\Delta P\times D}{4t\times E}

Applying values we get

\therefore \epsilon _{axial}=\frac{5\times 10^{6}\times 2.5}{4\times 20\times 10^{-3}\times 193\times 10^{9}}\\\\\therefore \epsilon _{diam}=\frac{2.5}{3088}

Hence The effect of axial strain along the diameter is given by

-\nu \epsilon _{axial}

Applying values we get

-\nu \epsilon _{axial}=-0.27\times \frac{2.5}{3088}=-0.0002185

hence

\epsilon _{D} =\frac{5}{3088}-0.0002185\\\\\epsilon =0.00140

Now by definition of strain we have

\epsilon _{D} =\frac{D_{f}-D_{i}}{D_{i}}\\\\\therefore D_{f}=D_{i}+\epsilon D_{i}\\\\D_{f}=2.5+0.0014\times 2.5\\\\\therefore D_{f}=2503.5mm

Increase in the diameter is thus 3.5 mm

Using the same procedure for axial strain we have

\epsilon_{axial} =\epsilon_{axial}-\nu \epsilon _{diam}

Applying values we get

\epsilon_{axial} =\frac{2.5}{3088}-0.27\times \frac{5}{3088}

\epsilon_{axial} =0.0003724

Now by definition of strain we have

\epsilon _{axial} =\frac{L_{f}-L_{i}}{L_{i}}\\\\\therefore \Delta L=0.0003724L_{i}

where L_{i} is the initial length of the cylinder.

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4 years ago
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