Recent content by amd123

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    Solving a Plane's Delivery Challenge: Dropping the Package at the Right Spot!

    That's it, I'm putting down what I originally had as my answer: "Plane's do not drop packages. If this plane dropped the package, then the package would be destroyed." But, we could assume the package is dropped from an altitude of 900m?
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    Solving a Plane's Delivery Challenge: Dropping the Package at the Right Spot!

    That's what I had thought as well, that the height of the plane was fixed. If i go with this approach, then the plane doesn't have two velocity components? Only one because the height is unchanging. Which would mean the package's velocity in the y component is due to gravity; it would be in...
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    Solving a Plane's Delivery Challenge: Dropping the Package at the Right Spot!

    i think so, but originally i had thought the plane is constantly at an altitude of 900m
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    Solving a Plane's Delivery Challenge: Dropping the Package at the Right Spot!

    I see what you are saying, in the initial velocity it gt is not involved, but AFTER it is dropped, yes. I think I have worked out a solution. Bare with me. 1. Vpack (y component) = Vplane*sinθ + gt 2. ((Vpack y)^2 - (Vpack nought y)^2)/(2g) = dy (altitude from which the package has to...
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    Solving a Plane's Delivery Challenge: Dropping the Package at the Right Spot!

    @lewando, the x-axis would be the distance's between the nose of the plane and the rear of the ship. The y-axis would be the difference in altitudes. Okay, I drew a diagram the velocity of the plane is given but it has an x and v component. Vplane = 180 m/s Thus the intial velocity...
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    Solving a Plane's Delivery Challenge: Dropping the Package at the Right Spot!

    Homework Statement A plane flies at 900m and has to deliver a package. It flies at 180 m/s North at 15° below the horizontal. The ships velocity is 40 m/s and it is traveling due north. At what horizontal distance from the ship must the package be dropped for it to land on the ship?Homework...
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    Deriving the Equation for Period of Simple Pendulum? Is my attempt correct?

    I have not come across any of the terminology you have just said. I doubt I would be allowed to use that method. Could you describe it in another way or comment on my attempt?
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    Deriving the Equation for Period of Simple Pendulum? Is my attempt correct?

    Homework Statement I have to derive an equation with the following variables: T= period m= mass Theta= angle l= length and I was told to think of the period as the perimeter of a circle.Homework Equations The Attempt at a Solution C=2\pi r,C/2=\pi r,d=\pi r,(at^2)/2=\pi r,t=\sqrt{\frac{2\pi...
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    Pulley but the cable is at an angle?

    In the diagram above, m1 = 285g, m2 = 755g, theta = 10.0 degrees, and mu = .047. If system is released from rest, how long does it take m2 to reach the end of table. Assum m1 can fall atleast 2.50 m.
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    Pulley but the cable is at an angle?

    I wish my PhD prof would realize this
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    Pulley but the cable is at an angle?

    http://img7.imageshack.us/img7/3597/doc092911.png Alright, that is how I'm setting up the problem.
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    Pulley but the cable is at an angle?

    The tension (resultant and tension for the hanging box) was calculated using the weight of the hanging box. Also, I'm in a Trig based physics course and we CAN'T do derivatives, actually we're not supposed to.
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    Pulley but the cable is at an angle?

    I meant to say that the tension of the falling box is equal to the resultant tension of the sliding box. The tension in the x direction would be cos (10) * Tension of falling box.
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    Pulley but the cable is at an angle?

    Homework Statement http://img813.imageshack.us/img813/594/75309500.png http://img813.imageshack.us/img813/7166/51848816.png I have to find acceleration to calculate how long it takes the heavier box to reach the end of the table. Homework Equations The Attempt at a Solution...
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