Recent content by EmittingLight

You dismissed my unpowered ball test in order to argue that a gravitationally accelerated frame was inertial, but then you used the same test to argue that the rocket accelerated frame wasn't. The equivalence principle is exactly why I changed the gravitationally accelerated problem to a rocket...

The point isn't about what we can and can't do though; the ball has a force acting on it, so it no longer has to have a straight line trajectory according to an inertial reference frame, in fact I think it can not possibly have a straight line trajectory (it only has one in a non-inertial...

I don't think that a frame being accelerated by gravity constitutes an inertial reference frame, and I also think that this means that there can not possibly be an inertial frame in our universe, at least not according to classical physics (i.e. gravitational force acts instantaneously...

Erm I just thought I'd clarify what the definition of an inertial reference frame is, since the OP has used terms/reasoning which implies that they haven't got a strong grasp on it. From the Wikipedia article: "An inertial frame of reference is one in which the motion of a particle not subject...

Well, didn't you say that the true answer was 42m before? It looks like it was rounded up from 41.67m to me. Do you know why you got t=-17.5 and d=-610 before? And not now? Based on what I've deduced, you actually already put in the correct values before. EDIT: I just thought I should add that...

Three of the values are known, I've told you what they are in all my previous posts: v_f, v_i and a. I'll put these into the equations for you:0^2=25^2+2*-7.5*d0=25+-7.5*tDo you know how to solve for t and d now? Remember, they are all in meters and seconds now, since that's what I've converted...

It's better if you show your working so that we can show where you went wrong. I suspect it might be unit conversions, which you can check with tools like Wolfram Alpha. The initial velocity is 90 km/h whereas the acceleration uses meters and seconds so you have to convert one of them to the...

Sorry I can't help you with this self-inductance business, but I do know that an ideal solenoid is one in which the magnetic field lines are completely straight parallel lines inside and outside of the solenoid. Real solenoids actually have bent magnetic field lines, which curve outwards when...

Well the kinematic equations can be found on this webpage. Then read the paragraph that follows the equations, and try to understand them well because you'll need to use them a lot. These equations ONLY apply when acceleration is constant. For this question you are told the: Initial velocity...

What about the friction? Imagine they put the brick on your arm. You would feel a downward force as you have calculated, yes. But if they started pulling the brick, would you feel your arm being pulled too, assuming that there was significant friction?

What do you mean by 'integrated 1 time'? Integrating a(s) from s=0 to s=1.3 is not going to give a velocity, the units will be \frac{m}{s^2}*m = \frac{m^2}{s^2}, while velocity has units of \frac{m}{s}. I'm not actually sure what grindfreak means when he says \int vdv=\int adx then "integrate...

But the acceleration is a function of the displacement such that if the displacement is zero, then the acceleration is also zero. It also happens that the velocity is zero initially, so the displacement would never change from zero and hence the acceleration would also never change from zero...

What I did there is called an indefinite integral. There are no limits, and is also the reason why I had to add a constant to the answer (I had to add one to both sides of the equation but I shifted both constants to one side to give the single c; another subtlety you might have noticed was that...

0.5=\frac{1}{2} So: a=-0.5v(t)^2/m=-\frac{1}{2}\frac{v(t)^2}{m}

Hmm...if a(s)=k*s*n and the car starts at s=0 with v=0, then the car will have an acceleration of 0 and will just stay stationary. Am I missing anything?