why is physics implied? aren't we talking about mathematical properties of an integral transform? this is something I can't understand.
Then, I still don't get if the lower bound is on uncertainties about bandwidth, duration or if it is about the magnitude itself of this quantities.
thanks! however this example only justifies that one spread or the other has to be non infinite...
in particular I'd like to have a somehow intuitive explanation of the fact that uncertainty principle gives a lower bound and not an upper bound..
Could you explain a bit about the relationship between locality and uncertainty in Fourier pairs?
Many pages talk about uncertainty principle stating that the precision at which we can measure time duration of signal cannot unlimitedly grow without affecting precision on bandwidth.
Many other...
Thanks for the correction! The fact that if passenger chooses the first wagon, then it doesn't choose the others made me think about "sampling without replacement"...
Mhh how could I approximate? I mean, I know that binomial can be approximated with normals.. but here I have a sort of hypergeometric multivariate distribution, isn't it?
I think this exercise exceed my course level.. I don't know how to proceed :(
Thanks a lot for the reply!
I thought of writing the number of interesting combinations as
\binom{150}{n_1} \binom{150-n_1}{n_2} \binom{150-n_1-n_2}{n_3} \binom{150-n_1-n_2-n_3}{n_4} \binom{150-n_1-n_2-n_3 -n_4}{n_5}
where n_1, n_2, n_3, n_4, n_5 < N and n_1 + n_2 +n_3 + n_4 + n_5=150...
Homework Statement
A train has got five train cars, each one with N seats. There are 150 passengers who randomly choose one of the cars. What is the probability that everyone will get a seat?
I think that what is asking me is "what is the probability that each wagon is chosen by no more than...
OOk, thanks.. you're right as always! :smile:
I also tried writing the momentum in the general frame:
M(U)+2M(U+ω L/2 cosθ)=2M ω0 L/2 (px at start)
so calculating for θ=0 I get
M(U)+2M(U-ω0 L/2 cosθ)=M ω0 L
U=1/3 (ω0) (L cos (θ)+ L)
... which gives Umax= 2/3 ω0 L that we got also with the use...
Uhmm, I think that the I that appears in rotational energy depends on axis of rotation.. isn't it?
On my notes I read:
## KE = \sum 1/2m_i v_i^2 = \sum 1/2 m_i r_i^2 \omega^2 = 1/2 (\sum m_i r_i^2) \omega^2= 1/2 I \omega^2##
Ohhh, I'm happy!
but I tried writing the conservation of energy, without using special frames, between t=0 and the position in which the rod has come back in vertical position:
##1/2 I \omega_0^2 = 1/2 I \omega_{\theta=0}^2 +1/2 M U^2+1/2 (2M) v_{cm}^2##/
but if ##\omega_0=-\omega_{\theta=0}##...
@TSny:
I corrected the calculation of Umax:
##U_{max} = -2/3 ω_{θ=0} L/2 cosθ + w =1/3 ω_{0} L + ω_0 L/3=2/3 L ω_0##
About PE:
At first (##\theta=0##), instead the motion is entirely rotational: ##E_{\theta=0}=\frac{1}{2}I\omega_0^2##, where I is the moment of inertia, calculated as...
I tried: (where A is friction force)
## F-A-T=M a_1 = M \alpha_1 2R ##
## T-A=M a_2 = M \alpha_2 2R ##
Subtracting, I get: ##F-2T=M(2R)(\alpha_1-\alpha_2)##
##F-2T=M(2R)\frac{3RF-2RT}{I}##
##T=F\frac{1+6MR^2 / I}{2+4MR^2 / I}##
Is it correct?
What about the work? Is it correct my reasoning in...