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Homework Help: Guys you gotta help me

  1. Sep 26, 2006 #1

    a solid of density 5000 kg m-3(m raised to 3) weighs 0.5 kg in air
    it is completely immersed in a a liquid in a liqiud of density 8000 kg m-3(m raised to three) .
    calculate apparent weight of solid in the liqiud

    plz help as soon as possible
    i gotta do ma homework

    P.S help meeeeeeee:eek: :uhh:
  2. jcsd
  3. Sep 26, 2006 #2
    hmm....weight is suppose to be in newton but urs is in kilogram??
  4. Sep 26, 2006 #3
    I need help with graphing!

    Um.. can you guys help me? I need to know what the graph would look like if I'm graphing the data for a ball droping downwards...
  5. Sep 26, 2006 #4
    Would the slope be going down and would the heights be negative on the graph?
  6. Sep 26, 2006 #5
    Weight is often used synonymously with mass so obviously he is meaning the weight of the object to be the mass of the object. Making the distinction between those two terms is a bit pedantic...

    AZ_Physics could you maybe shed some light on what information you have at your disposal? I need to know what you are plotting on each axis. Is it speed as a function of the distance from the point of release?
    Last edited: Sep 26, 2006
  7. Sep 26, 2006 #6


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    I hope you're not for real, since making the distinction between the terms 'weight' and 'mass' is of essential meaning for anyone who even wants to deal with the letter 'p' of the word 'physics'. :smile:
  8. Sep 26, 2006 #7
    Yes I am for real radou. I probably said it the wrong way though. I meant it's pedantic here in the 'intro to physics'. It's not to say that it isn't important to make the distinction, but when you're dealing with people who do a physics course as just an 'extra' I don't see it as the main point that should be made.

    I mean that statement really came across as a smartarse statement. It just said that weight should be in newtons. It didn't provide him any help or clearly point out that mass is different to weight. So I really wanted to point out that he wasn't being helpful at all.
  9. Sep 26, 2006 #8


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    I misunderstood the comments. Forget it.
    Last edited by a moderator: Sep 26, 2006
  10. Sep 26, 2006 #9
    :blushing: :blushing: :blushing: oops my bad.

    I completely misread that question. I thought it was an issue of finding the mass of the submerged part of the object.But it says it is completely submerged. I'm not thinking straight at all tonight. This is pathetic haha.
    Last edited: Sep 26, 2006
  11. Sep 27, 2006 #10
    if you are talking about the v-t graph, assume air resistance is neligible when falling in air, the graph would be y=x until a certain t before it becomes a straight line.for the object falling in liquid, it will be something like a log graph but tends to terminal velocity faster. hope that this helps.
    i was trying to know what his 0.5 is referring to................:frown:
    Last edited: Sep 27, 2006
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