Hello cnputhur, welcome to PF.
Your solution to the first question is the correct magnitude. However your minus signs may have gone awry. Try labeling the forces in future to help with these things ( a diagram will then make it very clear which way it must go).
And for question 2?...
Mathematically I would argue it something like:
The total energy of the system must be conserved, therefore
\textrm{K.E} + \textrm{P.E}= \textrm{const.}
\Rightarrow \frac{1}{2}mv(x)^{2} + mgh(x)= \textrm{const.}
The constant is arbitrary since we can define the potential as we...
You have worded this question somewhat strangely but I will attempt a solution. Think of the particle under SHM as a Pendulum clock (or weight on the end of a string). If it is "ticking" then ask "what is the velocity" at these two points:
1) When it is at maximum displacement from the vertical...
Hi,
when you say a force of 235Kg this does not make any sense. A Kg is a Mass, so under the acceleration of gravity it will have a force 235g=2350N in the negative vertical direction.
Since your diagrams do not provide this direction I can't tell you if you are making a mistake...
Hello mathuria1986,
This could be quite complex to solve, or simple depending on how this system is orientated with respect to the local gravity.
The 300Kg mass will have a force F=ma along this local gravity. You have drawn it such that there is an angle 23' to the x-axis (call it the...
I have no idea about a maximum number of votes, but who cares anyway!
Proportions are okay if you think about them in a good way (as with most things). Personally I like to introduce a constant so the statement "T is directly proportional to (r)^3/2" I would write as
T \propto r^{3/2}...
Hello KellySierra22,
Instead of giving your workings with numbers, it is easier to understand the problem in algebra.
We start with the well known F=ma, however we must be careful about which force and which mass (although the acceleration is constant). So I...
You can calculate the electric field using Gauss's law. And E(r) = Q / 4πε r^2, I believe is from a single point charge. You will need to calculate it for a wire a good discussion is given hear
http://farside.ph.utexas.edu/teaching/302l/lectures/node26.html
You may use that F=qVB when there is no electric field and the magnetic field is perpendicular to the direction of motion. To understand this consider that
\overline{v} \times \overline{B}=|\overline{v}||\overline{B}| \sin( \theta) \hat{n}
Since the magnetic field will be a loops in the...
I don't really think an understanding of equipartition theorem is required further than what Wikipedia can provide. Just consider it a big statistical problem (you have no interactions) and should be okay.
Indeed you have spotted your own mistake, when writing out the component of the 3N force along x, try to imagine two vectors parallel to the y and x axis. The magnitude of these is what you are physically trying to find.
You have written that
F_{x,2}=(−3)⋅\cos(110)≈1.03
I try not to...
Hello octowilli,
Thank's for providing nice diagrams and explanations of your attempt, in particular well done for solving algebraically before finding a numerical answer.
However you have become slightly unstuck when calculating the forces along x, I fear it may be a...
Hi nomalware80, have you tried these questions at all? Where do you get stuck, have you tried chucking some coins in a glass of water to answer questions 1.
It sounds like you are very close:
The more straightforward F = qvB(r) is true when you don't have an electric field and the velocity is perpendicular to the magnetic field, however you must be careful in the direction as there are some minus signs in a cross product like that.
Rather...