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A couple of physics questions for my project

  1. Mar 25, 2005 #1
    Why do two objects when dropped on the floor make different sounds? (i.e, coin and piece of wood)

    What's a system? and please give me an example of it demonstrating Newton's First Law of Motion.

    How do I calculate the speed of an object that demonstates linear motion?
  2. jcsd
  3. Mar 25, 2005 #2
    Sound is an audible vibration. Different mediums vibrate at different frequencies, and each of these frequencies you hear as different sounds. Theres alot of details as to why a medium vibrates at a certain frequency but I'm guessing you dont need to know why.

    A system is a compilation of any objects involved in an interaction. A ball is sitting on the ground, it is in equilibrium because there is no net force on the ball. When an outside force (you) kick the ball, the ball is no longer in equilibrium and now has a force acted upon it. The ball is then slowed down by air resistance and friction, as well as any objects it my hit in flight. If you consider all these as a system, then the total amount of energy in that system remains constant throughout.

    Speed or velocity:

    [tex] Velocity = \frac{\Delta Distance}{\Delta Time} [/tex]

    For an instantaneous velocity the equation is:

    [tex] Velocity = \frac{dx(t)}{dt}. [/tex] where x(t) is the position function. Thsi is derived using calculus.
  4. Mar 25, 2005 #3


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    or you could form a graph to find instantaneous velocity if you find trouble using calculas...
  5. Mar 25, 2005 #4
    You would use calculus to find instantaneous velocity no matter which way you do it :)
  6. Mar 27, 2005 #5


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    umm..im sorry to say this to you, but NO, we don't need calculas to find the instantaneous velocity. Theres no need to confuse the kid mate.
  7. Mar 27, 2005 #6
    Give me an applicable method to find instantaneous velocity without using calculus
  8. Mar 28, 2005 #7
    slope of a tangent of a d-t graph
  9. Mar 28, 2005 #8
    d-t graphs arent always linear. you would model the data to a function, derive that function, evaluate at a point to obtain the slope (velocity).

    even when they are linear, its just a case when the calculus is simple enough to be done with no work
  10. Mar 28, 2005 #9
    even on a curve .. it can still be done. this is grade 11 physics .. draw a straight line touching only the point where you want the velocity and find that slope.

    http://www.bfafairfax.com/~pfeiffer/Physics/KInematics/uniformmotion.html [Broken]
    under "Analysis of Graphs" an example of a tangent whose slope you would find
    Last edited by a moderator: May 2, 2017
  11. Mar 28, 2005 #10
    drawing a tangent simplifies the first sentence of my first post

    finding the velocity from the slope of that tangent line is what the second line was addressing
  12. Mar 28, 2005 #11
    I'm just trying to point out that you do not need calculus to figure it out
  13. Mar 28, 2005 #12
    We're both right. Just arguing different points of the same thing. Your claim is an instance of mine.
  14. Mar 28, 2005 #13


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    Then you must mean "calculus" in an extremely general sense- you seem to be claiming that any time you find the slope of a tangent line, you are using calculus.
    I would not use the word in that way: certainly Fermat and DeCartes were finding slopes of tangent lines long before Newton and Leibniz developed the calculus.

    Here's how I found speed in highschool physics BEFORE I learned Calculus:
    Draw the "position vs time" graph. Hold a mirror on the graph at the point in question and slowly turn it until the reflection of the graph in the mirror appears to line up with the actual graph. Draw a line along the base of the mirror. That line is perpendicular to the tangent line and you can use "compass and straight edge" to find the perpendicular to it (i.e. find the tangent line) or just repeat the mirror process with the perpendicular line. Once you have the tangent line, choose two points on it and use "rise over run" to find its slope (and that is NOT using calculus).
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