Recent content by athymy

Oh yeah, sorry about that. And thank you so much for the help, it was very informative.

Alright so since the width length 5nm, λ≈5nm or 5x10^-6m So v=h/(5*10^(-15)*5x10^-6m)≈2.65*10^-14m/s Is that reasonable? It seems like that is an extremely slow speed...

Ahh, so the wavelength should be approximately the size of the slit right?

If I remember correctly, doesn't the wavelength have to be pretty small in order for the particle to diffract? So could I try different values for v and see what my λ turns out to be? I still don't understand what the slit width has to do with the question. Is it just there to show that the...

Quote from Wiki: "Diffraction refers to various phenomena which occur when a wave encounters an obstacle. In classical physics, the diffraction phenomenon is described as the apparent bending of waves around small obstacles and the spreading out of waves past small openings"

Homework Statement A molecule is 3 nanometers wide and weighs 5 x 10^-15kg. It is fired through a slit that is 5 nanometers wide. Approximately how slow does the molecule have to go so that it diffracts? Homework Equations I'm thinking that I could use De Borglie's wavelength equation...

Thanks for your help! :D

Ahh, so Earth is approx. 150,000,000 km from the sun so is that my r value?

Homework Statement What is the escape speed of a spaceship fired in space, far away from Earth but at the same distance from the sun as the Earth. Homework Equations V[SIZE="1"]e = (2GM/r)^0.5 The Attempt at a Solution The problem is that I don't know what the M or r is since it...

Does anyone have a clue on how to solve this problem because I'm totally lost...

Homework Statement Hey guys, I'm having a hard time understanding this whole waves chapter and my teacher just gave us a tricky problem... So you have a single slit and it accommodates 3-6 wave lengths across its diameter. Assume that the sound created when echoing off objects may be...

Alrighty! Thank you so much for all the help, I really appreciate it :)

60MJ/Kg*70Kg=4200MJ (?)

Oops, sorry about that. I need to lift my mass which is 70kg.

I used the gravitational potential formula V=-Gm/r and that gave me -60MJ/kg