This mathematical model describes the changes that occur in a sample of water as its temperature increases. Based on this model,
how do you expect the motion of the molecules in ice to compare with the motion of molecules in water vapor?
A. I expect the molecules in ice to move more quickly than the molecules in water vapor.
B. This model does not provide enough data to answer the question.
C. I expect the motion to be very similar.
D. I expect the molecules n water vapor to move more quickly than the molecules in ice.
A. I expect the molecules in ice to move more quickly than the molecules in water vapor.
B. This model does not provide enough data to answer the question.
C. I expect the motion to be very similar.
D. I expect the molecules n water vapor to move more quickly than the molecules in ice.
2 answers:
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Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-)
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s
Answer:
The wavelength is 3500 nm.
Explanation:
d=
n= 1
θ= 30°
λ= unknown
Solution:
d sinθ = nλ
λ =
λ = 3500 nm