Recent content by TheBeesKnees

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    Egg Drop Experiment tips

    I used to do egg drops for science olympiad with popsicle sticks, but I found that a cone structure usually worked better than the rectangular ones.
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    Adiabatic process/ work done

    Hmmm... I've been searching, and I found the formula Q= 3/2 nRdT Which gives me 1.5*5.5*8.31*280= 19196 Joules, which my text book says is the correct answer. Where did the 3/2 come from?
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    Adiabatic process/ work done

    And in this problem, since it's adiabatic, it does not exchange heat, so if the temperature is going to go down, it must do positive work. I understand this. I just don't understand how to apply this stuff about Cv and Cp. I am at a loss for what I should do mathematically and what all...
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    Adiabatic process/ work done

    I see now that I was wrong to assume pressure is constant, but I'm afraid I'm still a little lost with the symbols and stuff. What does PV^gamma mean? Is gamma an exponent? And if so, do I need to figure out what both P and V are on their own? And what constant do they equal? My text...
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    Adiabatic process/ work done

    Oh yeah, and the answer to part c must be the opposite sign but same magnitude of the answer to a.
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    Adiabatic process/ work done

    Homework Statement During an adiabatic process, the temperature of 5.50 moles of a monatomic idea gas drops from 495 C to 215 C. For this gas, find (a) the work it does, (b) the heat it exchanges with its surroundings, and (c) the change in its internal energy Homework Equations I know...
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    A blowhard.

    That doesn't make sense still.
  8. T

    A blowhard.

    I got the difference in pressure as being 1812 Pacals. so that equals 1000kg/m3 * 9.8 m/s * delta h ? the height is .1848 meters?
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    A blowhard.

    so do I use the equation twice? 101.3 kPa = Pressure + (.5*1.29*53*53) Pressure = 99488 Pascals and then plug that in using water's density as rho the next time?
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    A blowhard.

    When I assume that, I get a little over 10 meters... and that can't be right.
  11. T

    A blowhard.

    for a) i got the radius of straw r=.003 m area and 1.5 L is .0015 m^3. Divide cubic meters by area of straw, and got 53.1 m/s But for part b) How do you use Bernoulli's eqn? Do you assume the pressure at the top of the straw is appoxamately zero, since the air is being blown off?
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