Recent content by MichaelTam

But how can I find t?## \lambda = 3 ## because ## k = 2 \pi/3 = 2 \pi/\lambda##

I know so ## y = A sin({2*10^6 ({\pi / 3 -\pi t }) + \phi }) ## There still some unknown I cannot find, can anyone give me some hint please?

Thanks a lot!

So I get 129.9 micro Newton .

Theta is ##pi/2 - pi/6##

Sorry , these are the updates.

I use the equation find that ## F = q v B sin(theta) ## F = 0.01*300*5*10^-5*0.5 F =75 N I don’t know why my solution is not correct?

I found the system is coming up with a error, the solution can be include m now, I’m sorrry….

I use the equation of 1.Potential energy = N*m*g*(h_2-h_1) 2.Energy of the battery can be used = (1-20%)*U (1) = (2) Then I find N=N=0.8*U/(m*g*(h_2-h_1)) but how can I express m in terms of other variable?

‘That’ means the last exercise it mention m is equal to 1000kg, I think I doesn’t work in this exercise.

the last exercise tells me m=1000kg, but I don’t know if I can substitute that into this situation or not.

But I cannot express N as variable m

each wall has a mass of m but N is the sum of those mass.

N=0.8*U/(g*(h_2-h_1)), you may see it at the second post

My solution is incorrect, but I still can’t find out the error….