Yep sorry haha spoilers. I didn't actually make the question and also not my solution however I do agree with their solution...
Is that 0.685 radians or degrees? The ^c is a little confusing...
Your approach makes sense, and is the way I did it before looking at his solution, however it yields a different answer
This is their solution for the first bit:
Spoiler
Radians. That's the approach I initially tried but I didn't know that trig identity so got stuck and looked for another way.
Mine doesn't work (you can tell that angle's too low because it must be greater than 45 degrees). You can draw parabolas that have maximum points at (x,2) where x is greater than two. I've given the launch angle for one of those I think and so the pizza hits the wall just below the roof.
1). Will hit the walls during its descent, due to the rotation of the planet. If you bore through the axis of rotation, as suggested, this won't occur.
2).
Simple:
Spoiler
Will update with a more complex solution revolving more around polar co-ords soon.
Even simpler, although not very rigorous because my internet's breaking on me (making LaTeX rather difficult):
Even simpler, although not very rigorous because my internet's breaking on me (making LaTeX rather difficult):
Spoiler
Indeed, if the earth was hollow, with a thin shell surface, there'd be no net force on a mass inside. You're suggesting splitting the earth up into an infinite amount of thin spherical shells. As you fall the shells you've already past will contribute nothing to the force acting upon you, but the shells further inside the earth, ones you haven't past yet, will still exert a force as if the mass was concentrated at the center, given by F=−r2GMm. It's exactly the same idea, just a different way of thinking about it. But showing the net force from a shell you've already passed is not as easy as the first method, if you're trying to be rigorous. Show some working, I'd like to see it all.
Indeed, if the earth was hollow, with a thin shell surface, there'd be no net force on a mass inside. You're suggesting splitting the earth up into an infinite amount of thin spherical shells. As you fall the shells you've already past will contribute nothing to the force acting upon you, but the shells further inside the earth, ones you haven't past yet, will still exert a force as if the mass was concentrated at the center, given by F=−r2GMm. It's exactly the same idea, just a different way of thinking about it. But showing the net force from a shell you've already passed is not as easy as the first method, if you're trying to be rigorous. Show some working, I'd like to see it all.
You don't need an infinite number of shells- compress all the mass you've passed into one thin subshell and all the mass further in than you acts like a point mass.
You don't need an infinite number of shells- compress all the mass you've passed into one thin subshell and all the mass further in than you acts like a point mass.
Spoiler
That's exactly what I said, the mass you've passed, its entirely irrelevant if you consider an infinite amount of shells, which is the only plausible way you could model it. And indeed, the shells you're yet to pass will act as a point mass, like I said, following the same force expression. You could say it was zero by intuition, but there is a way using geometry I think.
That's exactly what I said, the mass you've passed, its entirely irrelevant if you consider an infinite amount of shells, which is the only plausible way you could model it. And indeed, the shells you're yet to pass will act as a point mass, like I said, following the same force expression. You could say it was zero by intuition, but there is a way using geometry I think.
Ah, okay. I thought you were saying that the proof that it was zero was complicated. You could always form some complicated integral I guess and actually sum all of the attractions (I guess, my maths is worse than my physics...). I think I've seen a geometric proof for the inside of a hollow charged sphere which is the same situation... I'll try and find it.
Ah, okay. I thought you were saying that the proof that it was zero was complicated. You could always form some complicated integral I guess and actually sum all of the attractions (I guess, my maths is worse than my physics...). I think I've seen a geometric proof for the inside of a hollow charged sphere which is the same situation... I'll try and find it.
I meant to say, the proof of that, to be rigorous, is more complex, then just the solution I posted, unless you just take assumptions. It's the same proof really though.
Ah, okay. I thought you were saying that the proof that it was zero was complicated. You could always form some complicated integral I guess and actually sum all of the attractions (I guess, my maths is worse than my physics...). I think I've seen a geometric proof for the inside of a hollow charged sphere which is the same situation... I'll try and find it.
Two, free to move, small spheres with charge q and 4q are placed a distance d apart. A third small charged sphere, also free to move, is placed so that all three charges remain in their positions. Find the location and charge of the third sphere.
Two, free to move, small spheres with charge q and 4q are placed a distance d apart. A third small charged sphere, also free to move, is placed so that all three charges remain in their positions. Find the location and charge of the third sphere.
Is there more than one possible answer? Just by inspection. Will attempt later, didn't sleep last night, think I'm going to die.
Was hoping someone would answer this, thread is dying.
Firstly, we know the all the spheres need to be collinear. Thus, if you place the third charge some where inbetween the other two, we can solve this easily. Let the distance from charge 1 to charge 3 be r. We know 0<r<d
Solution 7
The force on charge 1 given by charge 3, must be equal and opposite to the force given by charge 2 on charge 1.
Let the charge of the third sphere be Q.
Thus:
ker2qQ=−ked24q2
This simplifies to Q=−d24qr2
Using the same logic for the two forces on charge 3:
ked24q2=−ke(d−r)24qQ⇒Q=−d2q(d−r)2
Equating either equations.
d24qr2=d2q(d−r)2⇒4r2=(d−r)2⇒(3r−d)(r+d)=0
Seeing as the charge lies between the other two charges, r=d. Thus r=3d
Subbing this in for the first equation, we get Q=−94q
The third sphere lies 3d away from the sphere with charge q, in between the charge q and 4q. It is 32d away from the 4q charge. The charge of the third sphere is −94q
Still doesn't make sense though.. If it's directly between the two charges but one charge is 4x larger than the other then the forces aren't equal
Give me two minutes and expel me from my Physics degree please. I used 4q for both charges, I think a common saying occurs here, RTFQ. You can see both expressions are (4q)(4q)≡16q2. I'll fix it now. I really shouldn't stay up so late and post.
Give me two minutes and expel me from my Physics degree please. I used 4q for both charges, I think a common saying occurs here, RTFQ. You can see both expression are (4q)(4q)≡16q2. I'll fix it now. I really shouldn't stay up so late and post.