I tried what they said above and I was able to get some answers. Probably not enough to get an A, but maybe enough to get some points to add to the other problems. I do not think you are biased. The suggestions about using my notes and textbook will not work, as there were no notes involving...
Yes. The professor mentioned that calculus is a method of solving these kinds of problems and many others, but he said we are not supposed to use calculus whatsoever.
I mean, for PE, the only thing that shows up is mgh when I search it online. I also know that Centripetal Force and speed are related in the fact that the object must be moving at a certain speed, or else it would get pulled to the center, or it would go flying away and leave orbit. I found an...
1. Homework Statement
If a mass attached to a spring has motion given by the equation X(t) = 5(sin(3pi(t))), what is the equation for the acceleration of the spring? What is the angular speed of the spring,and what is its frequency and period? If the spring has a spring constant of 900 N/m...
1. Homework Statement
Calculate the potential energy and kinetic energy of the moon, as well as it's escape velocity. Give the moon's period and angular speed. Consider the Earth to have a mass of 6 x 10^24 kg, the moon to have a mass of 7 x 10^22 kg, and their center separation distance to be...
That's most likely what I will do. Thank you for the help! :) yeah, it is somewhat unusual..I'm not sure in what scenario I'd use it, but maybe it'll show up sometime in the fluids section.
Thank you for the rapid response! Our class is algebra based physics, and our professor does not want us using things like integrals and other pieces of calculus. Is there a way to determine it algebraically? Thanks!
1. Homework Statement
Express the gravitational force equation in terms of density.
2. Homework Equations
F(Gravity) = ((GravitationalConstant)(Mass1)(Mass2))/radius^2
Density = mass/volume
3. The Attempt at a Solution
Based on the original equation for gravitational force, there are two...
OHHHH! The angle at which they're acting to the beam is 90 degrees. Sin of 90 degrees is 1, which means I simply need to set them equal to something since the beam isn't moving. Isn't the net torque if the beam is balanced 0? If that was the case, I would get the answer that the radius equals...
Oh, I see what you're saying. The upward force is 300N, coming from the fulcrum. The downward force also has to be 300N, coming from the weights of both m1 and m2. Since m1 is 20kg, I would multiply gravity (our professor has us use 10) by 20. That would give me a force of 200N. Which means that...
I guess, yeah. I just always type exactly what my professor sends us. :/ and would the ratio change at all then? Since it remained at 5m? The equation for the disk would remain the same at 1/2mr^2, but the equation for the single point mass would also remain at mr^2...right?
Well, since I still don't have a specific mass, I may have to put it in terms of m, leaving m as a variable...but at the same time, would the radius become zero, since all mass is concentrated at one point instead of people spread across it? I'm attempting to visualize it, and I'm seeing a...
Ah, so the ratio would be 2:1 in favor of the ring. That makes sense to me, though I'm still slightly unsure about the "moment of inertia" concept, though I'm sure I can learn that eventually. What about the second half of the problem? "How would this ratio change of the ring were replaced with...
1. Homework Statement
What is the ratio of inertia of a 5 meter radius disk to that of a 5 meter radius ring? How would this ratio change if the ring were replaced with a since point mass a distance 5 meters from a pivot point?
2. Homework Equations
I = mr2
3. The Attempt at a Solution
The...
That's the problem, I don't know how to do either. I know that the angle is 0 degrees, so that makes finding torque a little bit easier, maybe... sin(0) is 0, which would mean that the torque wouldn't exist, unless I'm missing something there. Other than that, I know that the distance from the...
Because since the length of the beam is longer on the side of m2, it cannot be heavier. Otherwise, the beam would pull down. If the fulcrum was centered over the beam, they'd be even then...
I think that we're supposed to assume that the beam's mass is not enough to actually affect what's going on. I honestly don't know what the wedge pushing up with 300 N means. I figured that we'd have to have a value of force given with the masses hanging on the beam. I've never seen the wedge...
1. Homework Statement
In the picture, m1 is 20kg, and the wedge pushes up with a force of 300 N. If the length L1 is 2 meters, solve for m2 and L2.
2. Homework Equations
I'm not sure
3. The Attempt at a Solution
The picture shows some sort of beam balancing on a fulcrum that is not...
1. Homework Statement
Suppose that the momentum of a system is 8 kg m/s. What is the kinetic energy of the system if it has a mass of 10kg? If this system is brought to rest with a constant force in 5 seconds, what is the magnitude of this force?
2. Homework Equations
Momentum > p = (m)(v)...
My first guess would be momentum, but I am not quite sure how that would come into play here...Unless you're saying that it would also be moving at 20m/s...
1. Homework Statement
A 5kg mass moving at 20m/s collides with a stationary 10kg mass. If the 5kg mass comes to rest, how fast will the 10kg mass be moving?
M1 = 5kg
M2 = 10kg
V1 = 20m/s
2. Homework Equations
KE = 1/2mv2
3. The Attempt at a Solution
I am fairly confident in this, and...