The Rite of Spring was composed by
1 answer:
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Answer:
a) v = 2.4125 m / s , b) Em_{f} / Em₀ = 0.89
Explanation:
a) This is an inelastic crash problem, the system is made up of the four carriages, so the forces during the crash are internal and the moment is conserved
Initial
p₀ = m v₁ + 3 m v₂
Final
= (4 m) v
p₀ =p_{f}
m (v₁ + 3 v₂) = 4 m v
v = (v₁ +3 v₂) / 4
Let's calculate
v = (3.86 + 3 1.93) / 4
v = 2.4125 m / s
b) the initial mechanical energy is
Em₀ = K₁ + 3 K₂
Em₀ = ½ m v₁² + ½ 3m v₂²
The final mechanical energy
= K
Em_{f} = ½ 4 m v²
The fraction of energy lost is
Em_{f} / Em₀ = ½ 4m v² / ½ m (v₁² +3 v₂²)
Em_{f} / Em₀ = 4 v₂ / (v₁² + 3 v₂²)
Em_{f} / Em₀ = 4 2.4125² / (3.86² + 3 1.93²)
Em_{f} / em₀ = 23.28 / 26.07
Em_{f} / Em₀ = 0.89
Complete question:
Suppose the wavelength of the light is 550 nm . How much farther is it from the dot on the screen in the center of fringe E to the left slit than it is from the dot to the right slit? Fringe C is the central maximum.
Check the image uploaded.
Answer:
The difference is 1100 nm
Explanation:
A bright fringe creates constructive interference, the wavelength is always in a multiple form.
At center fringe, the difference is (550 nm)(0) = 0
For the first maximum, the difference is (550 nm)(1) = 550 nm
For the second maximum, the difference = (550 nm)(2) = 1100 nm
Thus, for nth maxima, the difference is (550 nm)(n)
From the image uploaded, C is located on the second maximum, therefore the difference is given as (550 nm)(2) = 1100 nm
Therefore, the dot on the screen in the center of fringe E to the left slit is 1100 nm
