Recent content by lebprince

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    A disk with a Spring around it

    ok so i would use 0.05 x 27 = 1.35 and F = mxa so a = f/m = 27/2.5 = 10.8
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    A disk with a Spring around it

    ok..so i can find the torque on the body using r x Fnet, now is r the distance the center of the disk has moved? so 0.038 x 27 = 1.026? and F = mxa so a = f/m = 27/2.5 = 10.8
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    A disk with a Spring around it

    so i can use F = mxa?
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    A disk with a Spring around it

    umm am assuming i have to find the angular speed cause it would one one way to find V, but i know the Lrot = IW but i don't have Lrot
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    A disk with a Spring around it

    ok since we have 4 masses ; m1r^2 + m2r^2 + m3r^2 + m4r^2 so 0.2268
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    A disk with a Spring around it

    ok so the moment of inertia of the object is 1/2Mr^2 so = 1/2*2.5*(0.05)^2 = 0.003125
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    A disk with a Spring around it

    Homework Statement A string is wrapped around a uniform disk of mass M = 2.5 kg and radius R = 0.05 m. (Recall that the moment of inertia of a uniform disk is (1/2)MR2.) Attached to the disk are four low-mass rods of radius b = 0.09 m, each with a small mass m = 0.7 kg at the end. The device...
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    Calculating Angular Momentum with Constant Torque: A Particle in Motion

    Homework Statement At t = 15 s, a particle has angular momentum <5, 8, -3> kg · m2/s relative to location A. A constant torque <12, -13, 20> N · m relative to location A acts on the particle. At t = 15.2 s, what is the angular momentum of the particle relative to location A? The Attempt...
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    Collision of Alpha particle with a Gold Nucleous

    Ok there is a question after part c asking me for the final momentum of the gold nucleus long after the interaction with the alpha particle. i got the answer for that which was 2 * p1i, p1i is the initial momentum of the alpha particle. so 2 * 1.427e-19. now for part C i tried using the momentum...
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    Collision of Alpha particle with a Gold Nucleous

    Homework Statement An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 9.5 MeV (9.5106 eV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart...
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    Calculating Final Momentum in Inelastic Space Rock Collision

    so if we have an inelastic collision what would i need to calculate DeltaEthermalA + DeltaEthermalB? Thanks
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    Calculating Final Momentum in Inelastic Space Rock Collision

    ok thanks i used Efsystem = Eisystem and figured it out. Thanks but when thermal energy gets involved it confuses me like this question suppose that in the collision some of the kinetic energy is converted into thermal energy of the two rocks, where Ethermal,1 + Ethermal,2 = 1.44106 J. What is...
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    Calculating Final Momentum in Inelastic Space Rock Collision

    Thanks i got that part right. can i get some help on this? Suppose the collision was elastic (that is, no change in kinetic energy and therefore no change in thermal or other internal energy of the rocks). In that case, after the collision, what is the kinetic energy of rock 2? i wanted...
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