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Math Help - 5 Question from Practice Exam, Please Help Me!

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    5 Question from Practice Exam, Please Help Me!

    The Practice Exam I was given contains answer, but getting the answer is the hard part.

    1. A rotating wheel requires 3.00 s to rotate 37 revolutions. At the end of 3.00 s the angular velocity will be 98.0 rad/s. What is the constant angular acceleration? (1 rev = 2pi rad)

    Ans:13.7 rad/s^2


    Im using the formula delta(theta) = (omega)(time) + 1/2at^2
    delta(theta)= (37*2pi)
    (omega)(time)= 98.0 rad/s * 3sec
    1/2at^2= 1/2a(3sec)^2
    I plug everything in, but i get -13.7rad/s^2. Am I just lucky I came close to negative answer or did i so something wrong?

    2. A meter stick can be balanced at the 49.7-cm mark when balanced on a fulcrum (pivot). When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick in grams?

    Ans: 139 gm

    Yea, I'm clueless

    3. Two spheres of equal volume 100 cm3 contain an inert gas at 0C and 1.00 atm pressure. They are joined by a small tube of negligible volume, allowing the gas to flow between the spheres. What common pressure will exist in the two spheres if one is raised to 100C while the other is kept at 0 C? Assume an ideal gas.

    Ans: 1.16 atm

    I try every way possible to manipulate Pv=nRT, but I can't get the answer

    4.A 5.00–g bullet moving with an initial speed of 400 m/s is fired horizontally along the positive x-axis, passing completely through a 1.00-kg block, which initially is at rest on a frictionless horizontal surface at x = 0.00 cm. The block is connected to a spring having a force constant of 900 N/m. If the block moves 5.00 cm to the right (x = 5.00 cm) along the positive x-axis after impact, what is the speed of the bullet when it emerges from the block?

    Ans:100 m/s

    From the formual (KE+PEg+PEs)i = (KE+PEg+PEs)f
    I'm got the formula: 1/2mv^2 = 1/2mv^2 + 1/2Kx^2
    I don't know what I am doing wrong.

    5. A thin spherical shell of mass 0.400 kg and diameter 0.200 m is filled with alcohol (rho = 806 kg/m3). It is released from rest at the bottom of a tank of water (rho = 1000 kg/m3). What is the acceleration of the alcohol spherical shell as it starts to rise to the surface?

    Ans:1.06 m/s^2


    Yea, another clueless one

    Thank you, for whoever helps me on this
    Last edited by Heavyarms2050; May 1st 2008 at 03:28 PM.
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