How are interference patterns made?
1 answer:
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Answer:
4.15 m/s
Explanation:
As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:
ΔE = ΔK + ΔU =0
If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:
h₀ = 1.09 m -0.21 m = 0.88 m
⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J
As Uf = 0, ΔU = Uf -U₀ = -352 J
As the swing starts from rest, K₀=0, so we can say:
ΔK = Kf = (1)
As ΔK = -ΔU ⇒ ΔK = 352 J (2)
From (1) and (2) we can solve for vf, as follows:
So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.
Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original so it has 8.8 cm with the same orientation as original and it is a virtual imagen.
Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:
1/p+1/q=1/f where p and q represents the distance to the mirror for the object and imagen, respectively. f is the focal length for the concave mirror.
replacing the values we obtain:
1/8.3+1/q=1/15.2
so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3
then q=-18.28 cm
The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2
We also add a picture to see the imagen formation for this case.
