Recent content by benca

oops, yes I've worked it out now. thanks

I've first method I tried was using f = v/λ to find the frequency, then E = hf to find the energy and then using E = (-13.6eV/n^2) - (-13.6eV/n^2) to rearrange and solve for the unknown n. However I got 5, the same as the original entry level. I also tried using 1/λ = R(1/4 - 1/n^2) to solve...

Alright, thanks

I thought of that too, but I wouldn't exactly get Planck's constant but instead a number close to Planck's constant. But if that's what I need to do then that's ok, thanks

Attempt: I was thinking of finding the slope of the graph but I only know the values for x = 10, y = 3 and y = 0. And without the y-intercept, I don't know the work function and can't solve for h. If you can't see from the picture, the last co-ordinate is (10,3) and the x-axis is measured in f...

So I used dsinθn= (n - 0.5)λ to solve for the first two nodal lines and got 13.8° for the first and 46° for the second. When I tried to to input 3 for n I got a domain error. Does that mean the third nodal line is greater than 90° and didn't strike the screen? I also still don't know the reason...

I'm having trouble understanding what it's asking me. "Calculate the angles at which the nodal lines in the pattern are located far from the sources." I assume they are very far away, making lines PnS1 and PnC parallel. Is the question asking me to calculate θ' in the example? "nodal lines"...

Well in examples provided the satellites were much less. I guess I thought it should have been similar. And now that I say it, the direction of acceleration is towards the centre of the Earth (?)

a) Eg = Gme/r^2 r = √Gme/Eg r = √[(6.67x10^-11 N*m^2*kg^2)(5.98x10^24 kg)]/(4.5 N/kg) r = 9.41x10^6 m h = r2 - r1 h = 9.41x10^6 m - 6.38x10^6 m h = 3.03x10^6 m that's over 3000 km. Did I not use for right equation? Is Eg not 4.5 N/kg? Also for b), isn't the force of gravity the centripetal...

It's from an adult high school course I'm taking. There are no formal lectures, just small lessons I need to hand in. I don't know about other adult learning centres but I found out that this one is notorious for it's poor material quality, almost every lesson so far has had either outright...

a) F = J/t F = 4000 N*s / 0.35 s F = 11429 N b) I was going to equate impulse to the change in momentum and solve for v' (final velocity). Then use v' to solve for ΔEk. set ΔEk = Fd and solve for d. (The question never mentioned an angle of inclination, so I thought it would be ok to use W =...

oh my, I think I'm going to have to get used to odd numbers in this course. Thanks

Also, if there is no acceleration in the x direction, does that mean Fx+Ff=0 ?

Alright, I'm still wrapping my head around the idea of two normal forces and the difference between a downwards normal force and the force of gravity. I think I need to go over some material again. This actually does make a lot of sense, thanks for clearing that up. So, Fn+Fy+Fg=0 Fy=-Fn-Fg...

I went over it again and I think I have been consistent, at least in the y direction. Set right/up to positive and left/down to negative. The work given is positive so that means it's going in the same direction as the motion of the cart (to the right) Fn + Fay = Fg The numbers I substituted...