Firstly, do you understand the concept of a wavelength?
Answer:
Explanation:
Answer:
Explanation:
Given that,
System of two particle
Ball A has mass
Ma = m
Ball A is moving to the right (positive x axis) with velocity of
Va = 2v •i
Ball B has a mass
Mb = 3m
Ball B is moving to left (negative x axis) with a velocity of
Vb = -v •i
Velocity of centre of mass Vcm?
Velocity of centre of mass can be calculated using
Vcm = 1/M ΣMi•Vi
Where M is sum of mass
M = M1 + M2 + M3 +...
Therefore,
Vcm=[1/(Ma + Mb)] × (Ma•Va +Mb•Vb
Rearranging for better understanding
Vcm = (Ma•Va + Mb•Vb) / ( Ma + Mb)
Vcm = (m•2v + 3m•-v) / (m + 3m)
Vcm = (2mv — 3mv) / 4m
Vcm = —mv / 4m
Vcm = —v / 4
Vcm = —¼V •i
I need to see the diagram
Answer and explanation:
A.
Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)
B.
In muon rest frame it travels Zero meters
C.
Distance, d = Velocity, v * Time, s


D.
Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame
that is same as in part C which is 594m
E.
Using lorentz contraction
In the rest frame of someone standing on the mountain
the distance is given by




d' = 258.92m
F.
in the rest frame of someone standing on the mountain,
muon moves straight down
(1,500 meters) x (1 sec/330 meters) =
(1,500 / 330) (meters-sec/meters) =
4.55 seconds