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In case the question concerned the case $r \rightarrow 0$, you would reach the situation where the charge (represented by a charged particle like electron, proton, positron) approaches the Coulomb field of the other particle and they would have a tendency to create a kind of a planetary system - but - quantum effects start to play a role here and those two ...


3

If r = 0 then you have a single charge, so the problem reduces to the electromagnetic self-force problem. A charge will interact with the electric field it is in, and that includes the field due to its own charge.As long as the charge is not accelerating, one can pretend as if there is no self-force, but for accelerating charges, the self-force will lead to ...


2

You must solve for the objects initial velocity first: $$ v(t)=u+at\\ v(0)=u\\ v(1)=u+10\text{m/s}\\ =20\text{m/s}\\ u=10\text{m/s} $$ With this adjustment you should find the correct answer.


4

If the object was released from rest ($u=0\,{\rm m/s}$), what is its speed after $1\,{\rm s}$ if $a=10\,{\rm m/s}^2$? Using: $$v=u+at$$ you will find that the object was not released from rest...


2

You are correct. If you worked through the same steps on the distance ladder in a universe with a different speed of light, you would find the answer in light years would be different. Your question on whether light travels one light year in one year, though trivial sounding, opens the door to difficulties that are inherent in the nature of space-time. Your ...


1

force = acceleration * mass, hence acceleration will be $a=50N/22kg \approx 2.27m/s^2$ Distance it moves might be found by integration: $\int_0^{1.2}v(t)dt=\int_0^{1.2}atdt$, since speed $v(t)=at$ Answer to (B) then is 1/2*1.2*(2.27*1.2)=1.63, which seems pretty close to what you have got



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