Recent content by Ginny Mac

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    How Do You Add Angular Frequencies in Wave Simulations?

    I am working on creating a simulation of beats. I need to start with: (where w is angular frequency), w1 + w2 = w1,2 then w1,2 + w3 = w1,2,3 then w1,2,3 + w4 = w1,2,3,4 and so on, to create a wave packet where I will plot w1,2,3,4...n. My problem is that I do not understand how to...
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    Transverse Wave oscillator Problem

    If you have a moment, would you mind reviewing this? I appreciate it. Here is my dilemma this semester : I am confused as to how I ended up in a Calc-based Physics class (eek!) but I am trudging along best I can! Thank you very much for your help - Ginny
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    Transverse Wave oscillator Problem

    Here is a transverse wave problem: Consider transverse waves moving along a stretched spring fixed to an oscillator. Suppose the mass of the spring is 10 grams, its relaxed (not stretched) length is 1.00 meters, and its spring constant (if stretched) is 5.0 N/cm. The spring is tensioned via...
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    Why Is the Mass of the Equal-Arm Balance Neglected in Torque Calculations?

    In my lab assignment, we used an equal-arm balance to demonstrate rotational equilibrium by balancing torques using different mass quantities. A 100 g mass hangs from one end of the EAB, and a 185.18 g mass hangs from a thread at a 50 degree angle on the other end. The EAB is horizontal and we...
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    How to Calculate Momentum and Impulse in a Car Collision

    Hey ya'll - I am not getting the kind of answers I need to be getting for this lab assignment! Let's say that I am analyzing a car collision, and I have initial and final velocities (initial=before collision, final=after collision) and the masses of both cars. To calculate initial and...
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    Work required to pull a hanging chain

    Alright, thanks Doc! Check this out: (1) Center of Mass Method m=0.012 kg/.28 m = .0428kg/m*.07m = 0.003 kg. We'll call this m1, as it is the mass of the 0.07 m hanging off the table. Then change in U = (0.003 kg)(9.8 m/s^2)(.07m/2) <----(center of mass) =.001029 change in U =...
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    Finding the Equilibrium Separation in a Diatomic Molecule: A Scientific Approach

    Ahhh...so I got my equation down to (2A/B)^(1/6) = r. So that value of r is the equilibrium separation. Now for the next part, figuring out atomic behaviors based on larger/smaller values in relation to equilibrium sep. This should be interesting. Equation: 2A-Br^6=0 If sep. is larger...
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    Finding the Equilibrium Separation in a Diatomic Molecule: A Scientific Approach

    okay...but there are three unknowns in my equation... Is there a step I am leaving out?? Even solving algebraically only reduces down to 2A-Br^6 = 0. Does something cancel out that I am not aware of? Or maybe we are supposed to be left with an equation?
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    Work required to pull a hanging chain

    So if use m1x1 + m2x2 + m3x3 + etc.../M And plug in (.0429kg)(.07m) + (.0429kg)(.06m) +...etc. divided by 0.012 kg, = 0.99167m This will give me the location of the center of mass...and then I can use this to calculate the work?? I don't know where to go from here. What does this...
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    Finding the Equilibrium Separation in a Diatomic Molecule: A Scientific Approach

    Oh boy...this problem has stumped me. I am not even sure I know where to start: The potential energy of a diatomic molecule (a two-atom system like H2 or O2) is given by U= (A/r^12) - (B/r^6) where r is the separation of the two atoms of the molecule and A and B...
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    Work required to pull a hanging chain

    Hmmm..do you suggest that I calculate work for each little piece of chain, and then add them all together? For instance, (.012kg/.28m) = .0429kg/m. Then calculate for this mass at different distances, such as change in U = (.0429kg/m)(9.8m/s^2)(.07-0m) = .0294 J, then change in U =...
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    Work required to pull a hanging chain

    Okay...thanks Doc...I think I am beginning to understand. With that in mind, I used: change in potential energy = -W change in PE = mg(y-yinitial) change in U = (0.012 kg)(9.8 m/s^2)(0.07-0 m) change in U = .008232 J, or 8.2 * 10^-3 J So Work would be -8.2 * 10^-3 J. I...
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    Work required to pull a hanging chain

    A chain is held on a frictionless table with one-fourth of its length hanging over the edge. If the chain has length L= 28 cm and mass m=0.012 kg, how much work is required to pull the hanging part back onto the table? I have used this model: W horizontal + Work due to gravity = Work...
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    How Does Gravitational Force Affect a Snowball Fired from a Cliff?

    A 1.50 kg snowball is fired from a cliff 12.5 m high. The snowball's initial velocity is 14.0 m/s, directed at 41.0 degrees above the horizontal. (a) How much work is done by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in gravitational...
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    Struggling with Integration by Parts? Try a New Approach with Secant Functions!

    Integration by parts... I just started Calc. II and though I struggle a bit, it's fascinating. I have been fooling with a problem lately...one of those standard problems that professors like to assign, and it usually appears in calculus texts: Have ya'll ever done integration by parts with...
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