Recent content by TmrK

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    Finding out distance before breaking leg using Hooke's Law

    Depending on how you fall, you can break a bone easily. The severity of the break depends on how much energy the bone absorbs in the accident, and to evaluate this let us treat the bone as an ideal spring. The maximum applied force of compression that one man’s thighbone can endure without...
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    Multiple problems dealing with springs and Hooke's Law

    Homework Statement A vertical ideal spring is mounted on the floor and has a spring constant of 132 N/m. A 0.80-kg block is placed on the spring in two different ways. (a) In one case, the block is placed on the spring and not released until it rests stationary on the spring in its equilibrium...
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    Finding angle of rod hanging by springs

    Homework Statement A uniform 1.4-kg rod that is 0.60 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 59 N/m and 39 N/m. Find...
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    Finding the speed of a bullet via device with two rotating disks

    Homework Statement The device consists of two rotating disks, separated by a distance of d = 0.879 m, and rotating with an angular speed of 96.1 rad/s. The bullet first passes through the left disk and then through the right disk. It is found that the angular displacement between the two...
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    Finding the angle of the traveling direction of a sailboat

    So with this, W=F1(cosθ)d And plugging in the equations, F1(cos\Theta)55m = F2 X 42m (being that cos0°=1) divide F on both sides, cosθ55m = 42m divide 55m with 42m, and I assume that in the end, cosθ = .7(63 repeating)° cosθ = .76° Am I close?
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    Finding the angle of the traveling direction of a sailboat

    Homework Statement As a sailboat sails 55 m due north, a breeze exerts a constant force F1 on the boat's sails. This force is directed at an angle west of due north. A force F2 of the same magnitude directed due north would do the same amount of work on the sailboat over a distance of just 42...
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    Discovering mass of a star and orbital period of one of two planets

    So if I were to combine [2] and [3], I would get something like: M=v2\frac{vT/2\Pi}{G} or M=v3\frac{T}{2G\Pi}
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    Discovering mass of a star and orbital period of one of two planets

    Sorry, I'm really not used to using the "Show/Hide Latex Reference" button. I'll post the equations again. M=v2r/G T=2\Pir3/2/Square Root of GM Where G is the Universal gravitational constant, M is the mass of the sun, and T is the period (of the second planet).
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    Discovering mass of a star and orbital period of one of two planets

    Homework Statement Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be 41.7 km/s and 55.5 km/s. The slower planet's orbital period is 8.04 years. (a) What is the mass of the star? (b)...
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    Gravitational Force on three spheres

    Then if I really cannot use G again, then I would have to use all of the components to mind m3. However I can't being that there is no magnitude of the force mentioned at all. The problem alone I posted does not give me of what the magnitude of the forces are. Not one time is it even...
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    Gravitational Force on three spheres

    To find F, all I have to do is use G to replace as F, so really I'll be using G twice, yes?
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    Finding net gravitational force magnitude on moon

    Edit: did try finding it by vector sum, but did not worked as well. I'm not going to post what number I ended up with. It's that, or WileyPlus's system...
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    Gravitational Force on three spheres

    Exactly how does m3 cancel out? And also, I realized that the force all along was G itself. Please tell me I've gone one step forward in the positive direction.
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    Finding net gravitational force magnitude on moon

    Homework Statement The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force that the sun exerts on the moon is perpendicular to the force that the Earth exerts on the moon. The masses are: mass of sun=1.99 × 1030 kg, mass of earth=5.98 × 1024 kg...
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