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Oh, yes. My apologies, I missed that. Is r the initial radius, or is it one that changes as a function of time?
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Crap, I don't have my paper on me, but it gave a constant value for r and m1.
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Then I would suggest that you solve for g algebraically as Beary suggested. From there, I would determine the average value of g. While I know that this does not use a slope to get at the necessary value, I think that it should do the trick.
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Haha, that would have helped a ton! This problem has been bothering me and I've tried loads of ways to solve it(3 pages trying to work it out using simple harmonic motion).

If m1 and g are given and constant then just solve the equation as Beary said. You should get
g = (4π2*m1*r)/(m2*T2)

To get a linear graph for g, plot m2 and 1/T2
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Thanks, that's kinda what my teacher said, but he didn't explain it very well.

Could somebody explain why this is?
           Length   k
Spring 1   l        k
Spring 2   1/2*l    2*k

With the equation:

ForceApplied = k * Displacement

In other words, why a spring with half the lenth has twice the k?
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Yes. I think the peice of information that they are not telling you is that force is being kept constant.

Granted there is more to a spring than the simple equation F=-kd, but it works well enough for what you are doing.
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It doesn't explain why a shorter spring has twice the constant, though. Well, it kinda does, but not in any intelligible wording.
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I have a feeling that length is actually length of displacement, rather than the actual length of the spring. I believe that the length of the spring is factored into the spring constant.
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That's where I'm confused. It does give some explanation why the length of the spring (that is, the actual length, not the displacement) changes the constant k, but it doesn't make much sense. That's my question, is why it changes the constant.
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Imagine 2 springs, each of length L. For a given force, f, they will expand by a displacement of D.
Now join them end to end. For the same applied force, each one will expand the same distance, D, giving a total displacement of 2D. Now instead of thinking of them as two springs, think of them as one longer one.
The longer spring expands more for the same force applied. Alternatively put, a shorter spring needs more force to create the same displacement.

Intuitively this makes sense. Consider trying to stretch a small elastic band over a given distance, compared to a longer (but otherwise similar) band. You have to deform the small one more, which takes more effort.

Finally, please note that this rule is only approximate, due to the physical limits of RL springs.
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This, this makes sense. Thank you. I might be back with more questions. I have to teach myself a whole Physics chapter.
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I apologize in advance for the double post, but those who're most likely to help have already looked.

A 1.1 kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is stretched from its unstrained length.

Easy enough. ForceApplied = k * x, 1.1kg * 9.8m/s = 120N/m * x, x = .09m

Now, (b) The object is pulled straight down by an additional distance of 0.20m and released from rest. Find the speed with which the object passes through its original position on the way up.

I have no idea how to do this one. Everybody's help is always greatly appreciated, by the way. I think I'd fail this class without you all.
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If you know simple harmonic motion you can take y(t)= ACos(wt). A is the amplitude (the distance from equilibrium). Omega (I use w, because I don't have the symbol) is equal to Sqrt[k/m] wear k is the spring constant and m is the mass of the object.

So now, you can derive it to get your velocity equation. It will look something like: v(t)=-AwSin(wt). The maximum velocity is achieved when the spring is exactly at the equilibrium position. The maximum velocity occurs when Sin(wt)=1, so that means that the maximum velocity is Aw.

If you aren't familiar with the simple harmonic motion equations, I don't know what to tell you. I suppose you could derive them yourself.
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Is there an easy way to find the sum of the primes between a and b, where a < b?
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Brute force.
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On to Impulse.

A volleyball is spiked so that it's incoming velocity of +4.0 m/s is changed to an outgoing velocity of -21 m/s. The mass of the volleyball is 0.35 kh. What inpulse does the player apply to the ball?

Yeah, I'm guessing this is an easy question, but I have to teach it to myself. And I'm not even sure what it's asking.

Edit: Damn, no wonder I screwed up that post so much. My iPod keyboard was set on Spanish.
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Impulse is the change in momentum (or the integral of force over time).
Momentum is mass * velocity.

Change in velocity is from +4 to -21, or -25 m/s
Mass is 0.35
So the impulse is -8.75 Ns. (The - indicates a direction in the framework used to measure the velocity.)
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Is there a way to find the "length" of a number, without using logarithms?
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As in the order of magnitude?
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You could maybe use a series approximation of log(n), but that amounts to the same thing. Especially since a calculator is going to use an approximating function as well, implemented in hardware.
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Currents Question:

A metal bar, mass 0.04kg, resistance 4 Ohms and length 0.5 metres is placed on two frictionless conducting rails in a magnetic field of 0.3 Teslas. The bar is connected to a 36 Volt power source.

a) Calculate the current in the bar.

Well, the Current, I = V/R, I = 36/4

Therefore Current = 9 Amps.

b) Calculate the magnitude of the force which acts on the bar. F = BIl, F = 0.3 x 9 x 0.5

Therefore Force = 1.35 Newtons.

c) If the bar is free to move, and the force is in the appropriate direction, calculate how far it will move along the rails in 0.25 seconds.

[This is the one I'm stuck on.. Not sure how to use weight or time to find the force. We've only been learning about the formula F=BIl sin theta]
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You already calculated force in (b), didn't you? And you're given time, so you just need to use one of the kinematic equations to solve for distance.
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Yeah. F=m*a and s=.5*a*t2
So a=F/m=1.35/0.04=33.75
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Edit: GOt it through another source.
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The Forum > Math & Science > Math Problem Thread
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