Recent content by XxMuDvAyNexX

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    Another easy Conservation of Energy Question

    A girl swings back and forth on a swing with ropes that are 4.00 m long. The maximum height she reaches is 2 m above the ground. At the lowest point of the swing she is .5 m above the ground. (a) The girl attains the maximum speed (1)at the top, (2) in the middle, (3) at the bottom of the swing...
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    Force with kinetic energy question

    Ahh I think figured it out. The W would be equal to =-153.125J. Then I would just use F=W/D or F=-153.125 J/.12m and that comes out to be....-1276.04. Is that correct?
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    Force with kinetic energy question

    They are equal aren't they? W=K-Ko
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    Force with kinetic energy question

    [SOLVED] Force with kinetic energy question A 2.5-g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 12 cm into the tree's truck. What force was exerted on the bullet in bringing it to a rest? Not sure on this one. I know to find to...
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    Easy Spring Constant Question

    Ah man! I can't believe I missed that. Thank you!
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    Easy Spring Constant Question

    [SOLVED] Easy Spring Constant Question A particular spring has a force constant of 2.5 x 10^3 N/m. (a) How much work is done in stretching the relaxed spring by 6.0 cm? (b) How much more work is done in stretching the spring an additional 2.0 cm? I am using W=1/2kx^2 but I am not...
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    Newton's Second Law of Motion acceleration problem

    wait...Ah I'm doing another problem...sorry!! See I can figure out the problems that don't involve angles...those angles are so complicated! Ah back to trying to solve the first one... the one I solved was..."A jet catapult on an aircraft carrier accelerates a 2000 kg plane uniformly from rest...
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    Newton's Second Law of Motion acceleration problem

    I think I figured it out....First off... M=2000 KG Vi=0 Vo=320 km/hr = 88.88 m/sec T=2.0 sec Ok, I then used the equation D=1/2(Vo+Vi)T which turned out to be 88.88 M. So D=88.88 M. Then I used the equation A=V/T. That turned out to be 44.44 m/sec^2. Finally, I used Fnet=ma. This turned out to...
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    Parallel vs. Series HELP!

    In series the amps are the same throughout every resistor. In parallel the amps all add up to a total amperage. I'm pretty sure I know how to do it..see if you can figure out the rest.
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    Newton's Second Law of Motion acceleration problem

    An object (mass 10.0 kg) slides upward on a slippery vertical wall. A force F of 60 N acts at an angle of 60 degrees (the force is 60 degrees South of the x+ axis, it's shown in a picture) Determine the normal force exerted on the object by the wall. Next, determine the object's acceleration...
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    Angles and forces/normal force

    Man...sorry about all that. I can't seem to do any of these problems from this chapter without intense meltdowns or whatever you want to call it. I'm going to try from the beginning of this chapter and work my way through.
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    Angles and forces/normal force

    Oh I forgot to square 25. So...would it be 625/9.8? Which gives me 63.77! My book usually rounds this has to be it!!! Please tell me it is!
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    Angles and forces/normal force

    D=V/A? I've never seen that equation before. woah woah I'm getting closer...VF^2=VI^2+2A(D) so in other words D=VF^2-VI^2/2A.....VF=0 (since it's at rest)...VI would be 25 m/sec and A is 4.9 m/sec^2. So...I get arggggg!! I have a program that does it for me but it doesn't show what it's doing. I...
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    Angles and forces/normal force

    Ah sorry. I forgot that I cleared the RAM earlier. So it's 4.9 m/sec squared. What do I do next?
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