1. So you're saying h = gt^2 instead of 1/2gt^2? Well, you start with constant acceleration, g. Starting from rest, this means your velocity is gt. How do you get the distance traveled from the velocity? IF the velocity were constant, you would just take velocity * time. But your velocity is...
Yes, moments of inertia add. It is after all an integral of the form
\int r^2 \, dm
so you're free to break up an object into chunks, find the moment of inertia of each part and then sum them up altogether in the end.
yeah. alternatively, you should realize that the number 21.582N corresponds to the NET force on m_2. To find the horizontal force F, you've got to add the frictional force exerted by block 2 on block 1, which is 8.6328. You should get the same answer.
You're almost there. In your last step you used Newton's second law to find a force on m_2. But this force isn't actually the horizontal force F. What is it?
Ha is the set containing all elements of the form h*a, where h is in H.
What is the definition of a subgroup? It must non-empty, closed under multiplication and closed under inversion. So if Ha = H, then h_{1}a = h_{2}. What does this say about a?
If a is in H on the other hand, and H is...
Why don't you give it a try first: draw a force diagram, write down some equations. Remember the conditions for equilibrium: there should be no net force on the ladder, and no net torque about any point.
I assume the question is to find the Power of the system at the instant when a force of 2N is acting on a particle moving at 0.028 m/s. The power is work done divided by time, NOT kinetic energy of the particle divided by time. The kinetic energy of the particle is equal to the work that HAS...
Assuming the protons begin at rest, every second you have to deliver 1E9 protons at 11.3keV each. How much energy do you need to deliver per second, i.e. how much power?
If you had a constant force acting on the proton, then you would have a constant acceleration. You know what the final...
So a constant power is supplied. How much energy is supplied in a time t? Where does the energy from the power supplied go? If you know these, you should be able to get the required expression.
To eliminate c, since \eta = cr, you can integrate the charge density over the disk to compute the total charge, Q. This should give you c in terms of Q and R.
There are two isothermal and two adiabatic processes in the Carnot cycle. After the isothermal compression, we have an adiabatic compression to return to the original state.
You've got the right idea. You need an expression for dx in terms of dt, because dt is the measure of probability: the more time spent in that interval x and x+dx, the greater the probability of finding the particle at x.
You listed v = aw cos wt, but v = dx/dt. So this looks like a good...