The FREQUENCY of light remains unchanged once it leaves the source.
Answer:
The magnitude of the acceleration is 
The direction is
north of east
Explanation:
From the question we are told that
The force exerted by the wind is 
The force exerted by water is 
The mass of the boat(+ crew) is
Now Force is mathematically represented as

Now the acceleration towards the north is mathematically represented as

substituting values


Now the acceleration towards the east is mathematically represented as

substituting values


The resultant acceleration is

substituting values


The direction with reference from the north is evaluated as
Apply SOHCAHTOA

![\theta = tan ^{-1} [\frac{a_e}{a_n } ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%20%5E%7B-1%7D%20%5B%5Cfrac%7Ba_e%7D%7Ba_n%20%7D%20%5D)
substituting values
![\theta = tan ^{-1} [\frac{0.808}{1.269 } ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%20%5E%7B-1%7D%20%5B%5Cfrac%7B0.808%7D%7B1.269%20%7D%20%5D)
![\theta = tan ^{-1} [0.636 ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%20%5E%7B-1%7D%20%5B0.636%20%5D)

Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s
Answer:
Y, X, Z, W
Explanation:
You know W is the most recent because it features the nucleus in the middle and the electron cloud which was shown in models after the others.