Recent content by learnitall

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    Two Dimensional Motion (Seemingly Simple)

    Sorry for the late response. I had classes all day and got caught up with other hw. So I thought about what you said but instead of considering velocities I set up a simpler equation (after all this was an exam so there HAD to be easier way). I considered displacement: x(2)-x(1) = 13...
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    Two Dimensional Motion (Seemingly Simple)

    Homework Statement An object starts from rest at time t=0.00 s and moves in the +x direction with constant acceleration. The object travels 13.0 m from time t=1.00s to time t=2.00 s. What is the acceleration of the object? a) 5.20 m/s^2 b) 10.4 m/s^2 c) 8.67 m/s^2 d) 6.93 m/s^2 e) 12.1...
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    Which Iceboat Finishes with Greater Momentum?

    Ah, i see now. I didnt need that extra (1/2) factor. I also understand why I didnt need it. Thanks guys
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    Which Iceboat Finishes with Greater Momentum?

    Homework Statement Two iceboats hold a race on a frictionless horizontal lake. The two iceboats have masses m and 2m. Each iceboats has an identical sail, so the wind exerts the same constant force F on each boat. The two ice boats start from rest and cross the finish line a distance s away...
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    How Do You Calculate Mass M1 in a Frictionless Incline Problem?

    Think about Newton's second law: F=ma What is the acceleration of m1? What is the acceleration of m2? consider gravity and the coordinate system Mentz114 has chosen when drawing free body diagrams for both blocks. That should answer your question.
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    Distance traveled sliding down incline

    this makes a lot of sense now, thanks for the input!
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    Distance traveled sliding down incline

    You were right. I was missing the work don't by the parallel component of gravity which was +mgsin(theta)d. Using this quantity for the equation gives : d= .775 which is answer 'b'. Thank you Doc Al
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    Distance traveled sliding down incline

    I forgot to mention that I'm using g=9.8 m/s^2 for gravity
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    Distance traveled sliding down incline

    Ok so if I consider the work done by gravity's horizontal component through the distance 'd' as -mgdcos(theta) and plug that value into the left side of my Work-Energy equation. My new result is: d = .1087006786 m or approximately .108 m. My new result is off by a factor of 10 from the choice...
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    Distance traveled sliding down incline

    Homework Statement A box is sliding down an incline tilted at a 11.1° angle above the horizontal. The box is initially sliding down the incline at a speed of 1.70 m/s. The coefficient of kinetic friction between the box and the incline is 0.390. How far does the box slide down the incline...
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