#$&*
Phy 122
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bernoullis units conversion
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I'm working set 5 problem 5. The math is not a problem, but somehow I can't get the final units to N/m^2
The change in .5 `rho v^2 will be .5 `rho v2^2 - .5 `rho v1^2 = .5 `rho (v2^2-v1^2) = .5 * 1000 kg / m^3 * [ ( 8.5 m/s)^2 - ( 6.2 m/s)^2 ] = 16905 N/m^2.
The pressure change will be the negative of this, or -16905 N / m^2.
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It seems to me we have rho times velocity squared.
(kg/m^3)*(m^2/s^2)
When I cancel out I get (kg/m)*(1/s^2)
I beleive a Newton is kg * m/s^2. So, when we say N/m^2 we are saying (kg*m/s^2)/m^2
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Can you show me how the pressure units in Bernoulli's equation cancel to N/m^2 ?
I know I should be able to do this, but I'm having a mental block!
Rewriting m^2/s^2 as (m/s^2)*(m/1)I just tried (kg/m^3)*(m/s^2)*(m/1) and I can see how that could equal N/m^2. Is my rewrite correct? Thanks.
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rho v^2 has units of kg / m^3 * m^2 / s^2 = kg / (m s^2).
N / m^2 = (kg m / s^2) / m^2 = kg / (m s^2).
So it's clear that the two calculations have the same units.
To get from kg / m^3 * m^2 / s^2 to N / m^2 you could take kg m/s^2 out of the expression as follows:
kg / m^3 * m^2 / s^2 =
(kg * m / s^2) * m / m^3 =
kg m/s^2 / m^2 =
N / m^2.
rho g y is worth looking at as well, with units of
kg / m^3 * m/s^2 * m =
kg m/s^2 * (m / m^3) =
kg m/s^2 / m^2 =
N / m^2.
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#$&*
Phy 122
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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asst 6 lab Kinetic Model Exper
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I'm doing asst 6 lab Kinetic Model Experiment. I've read through the lab.
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I'm not clear on what I need to turn in for this lab. Should I just turn in the Research Questions or copy the entire thing?
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Just copy the questions and insert your answers.
The graphs don't show on the version of the program I have posted, so there are some questions you won't be able to answer.
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