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February 12th, 2019, 09:25 AM   #1
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Formula

Hi All,

Is there a formula to convert 1.21x10^8 1s-1 to 1.22x10^8kg (3 s.f)?

Thanks for any help.
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February 12th, 2019, 10:01 AM   #2
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there has to be more to this than you've written....

why should you be able to convert $Hz$ to $kg \cdot ft^2$ ?
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February 12th, 2019, 11:48 PM   #3
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Hi Romsek,

I am trying to carry out a calculation for an overspill of a damn.
When it overflows 1.21x10^8 1s-1 exits, how can I show that the mass of water leaving the lake is 1.22x10^8 kg?

Density - 1012 kg m-3
g = 9.81ms-2

Thanks.

Last edited by skipjack; February 13th, 2019 at 06:26 AM.
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February 13th, 2019, 05:41 AM   #4
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Quote:
Originally Posted by NAC54321 View Post
Hi Romsek,

I am trying to carry out a calculation for an overspill of a damn.
When it overflows 1.21x10^8 1s-1 exits, how can I show that the mass of water leaving the lake is 1.22x10^8 kg?

Density - 1012 kg m-3
g = 9.81ms-2

Thanks.
There is still a lot of confusion. How can 1.21 x 10^8 /s exit? That's a frequency, not an amount. Are you trying to say that water leaves the lake at a rate of 1.21 x 10^8 kg/s?

Can you please post the whole question?

-Dan
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Last edited by skipjack; February 13th, 2019 at 06:26 AM.
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February 13th, 2019, 05:57 AM   #5
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Quote:
Originally Posted by NAC54321 View Post
Hi Romsek,

I am trying to carry out a calculation for an overspill of a damn.
When it overflows 1.21x10^8 1s-1 exits, how can I show that the mass of water leaving the lake is 1.22x10^8 kg?

Density - 1012 kg m-3
g = 9.81ms-2
I'm assuming the units in bold are liters per second ... $Ls^{-1} = L/s$

$(1.21 \times 10^8 \, L/s) \, \cdot (10^{-3} \, m^3/L) \, \cdot (1012 \, kg/m^3) \, = 1.22 \times 10^8 \, kg/s$

... note the water leaving the lake is a rate of kg per second, not just kg.
Thanks from greg1313, topsquark and NAC54321

Last edited by skipjack; February 13th, 2019 at 06:25 AM.
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