Alan

The corresponding expression for a 2 bar applied pressure is:

$$\tau=\frac{\sqrt2A}{ag}(\sqrt{\frac{\Delta p}{\rho}+gh_0}-\sqrt{\frac{\Delta p}{\rho}})$$

Δp = 2 - 1 = 1 bar = 10^5 N/m^2  assuming 1 atmosphere = 1 bar

Assuming the liquid is water with a density of 1000 kg/m^3 then:

$${\mathtt{time}} = {\frac{\left({\frac{{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{8}}}}{{\mathtt{9.8}}}}\right){\mathtt{\,\times\,}}\left({\sqrt{{\frac{{{\mathtt{10}}}^{{\mathtt{5}}}}{{\mathtt{1\,000}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{10}}}}{\mathtt{\,-\,}}{\sqrt{{\frac{{{\mathtt{10}}}^{{\mathtt{5}}}}{{\mathtt{1\,000}}}}}}\right)}{\left({\mathtt{3\,600}}{\mathtt{\,\times\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{365}}\right)}} \Rightarrow {\mathtt{time}} = {\mathtt{1.862\: \!986\: \!880\: \!688\: \!817\: \!1}}$$

time ≈ 1.9 years

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A = pi*(0.25/2)^2  m^2

a = pi*(0.000025/2)^2  m^2

A/a = 10^8

g = 9.8 m/s^2

h0 = 10 m

so

$${\mathtt{time}} = {\frac{{{\mathtt{10}}}^{{\mathtt{8}}}{\mathtt{\,\times\,}}{\sqrt{{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{10}}}{{\mathtt{9.8}}}}}}}{\left({\mathtt{3\,600}}{\mathtt{\,\times\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{365}}\right)}} \Rightarrow {\mathtt{time}} = {\mathtt{4.529\: \!970\: \!283\: \!394\: \!941}}$$

time ≈ 4.5 years

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There is no solution to these. See the following for an explanation:

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If the numbers you list represent values of x then perhaps you are meant just to calculate corresponding values of y by substituting the numbes into the formula:

When x = 0  y = 2/3*0 - 1  so y = -1

When x = 3  y = 2/3*3 - 1 so y = 2 - 1  or y = 1

and so on, so you get a list of (x, y) pairs:  (0, -1), (3, 1) etc.

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3g + 8b = 46       (1)

8g + 3b = 31       (2)

Multiply (1) by 8 and (2) by 3:

24g + 64b = 368      (3)

24g + 9b = 93          (4)

Subtract 4) from (3)

55b = 275   

So b = 275/55 or b = 5

Put this back into (1)

3g + 40 = 46

3g = 6

g = 2.

You can now work out 2g + 3b

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Rearrange the first equation to get x = (y+2)/4 + 11

Put this into the second equation

3*(y+2)/4 + 3*11 + (y+2)/3 = 7

Multiply through by 4*3

9*(y+2) +12*3*11 + 4*(y+2) = 12*7

So

9y + 18 + 396 + 4y + 8 = 84

13y + 422 = 84

13y = -338

y = -26

Put this back into the first equation above

x = (-26+2)/4 + 11

x = 5

You should check these by putting them back into the original equations.

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(20 - 3)*(20 + 4) = 20^2 +20*4 - 20*3 - 3*4 = 400 + 20 - 12 = 408

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-log(a) is the same as log(1/a) where a is positive.

log(-a), where a is positive, is undefined in the domain of real numbers.

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In logic, "and" usually means multiply, "or" means add.

1 and 0 = 0   

1 or 0 = 1

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