Answer:
Δμ = hΔf/B
Explanation:
If the photon energy , ΔE = hΔf where Δf = small frequency shift and since the potential energy change of the magnetic dipole moment μ in magnetic field B from parallel to anti-parallel state is ΔU = ΔμB. where Δμ = small shift in magnetic moment.
Since the magnetic energy change equals the photon energy,
ΔE = ΔU
hΔf = ΔμB
Δμ = hΔf/B
Answer:
edieson and atela script once upon a time there was the name is a copy paste and the you can come by the office and I have a car so we can do it tomorrow if I will be working from the gathered that was a week if we were in w
Answer:
Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?
Explanation:
So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s

so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull
Horizontal speed = 24.0 m/s
height of the cliff = 51.0 m
For the initial vertical speed will are considering the vertical component. Therefore,
Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means

let's find how long the ball remained in the air.
![\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%200%3D51-%5Cfrac%7B1%7D%7B2%7D%289.8%29t%5E2%20%5C%5C%204.9t%5E2%3D51%20%5C%5C%20t%5E2%3D%5Cfrac%7B51%7D%7B4.9%7D%20%5C%5C%20t%5E2%3D10.4081632653%20%5C%5C%20t%3D%5Csqrt%5B%5D%7B10.4081632653%7D%20%5C%5C%20t%3D3.22%20%5C%5C%20t%3D3.22%5Ctext%7B%20s%7D%20%5Cend%7Bgathered%7D)
Finally, let's find the how far from the base of the building the ball landed(horizontal distance)
Regular reflection is a reflection such that the the angle of reflection of the light is equal to the angle of incidence and it is on the opposite of the normal to the point of incidence. In a regular reflection the incident ray, the reflected ray and the normal to the reflection all lie in the same plane. Regular reflection happens if the reflection surface is very smooth.