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katovenus [111]
3 years ago
8

How can you use the movement of molecules to determine when a phase change has occurred?

Physics
2 answers:
solong [7]3 years ago
6 0

Answer:

Explanation:

<em>These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.</em>

baherus [9]3 years ago
5 0

Answer:

These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.

Explanation:

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Mariana [72]

Answer;

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Explanation;

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6 0
3 years ago
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A(n) ______ is a gap in the geologic record where some rock layers have been lot because of erosion
fomenos

geological time is the answer

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3 years ago
In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the mea
Vikentia [17]
For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c). 

fringe = (delta t) / (λ/c) 

We can find (delta t) with the equation: 

delta t = [v^2(L1+L2)]/c^3 

Derivation of this formula can be found in your physics text book. From here we find (delta t): 

600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13 

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This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s. 
4 0
3 years ago
Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o
spin [16.1K]

Answer:

The true weight of the aluminium is m_{alu} = 4.5021 kg

Explanation:

Given data

m_{app} = 4.5 kg

\rho_{air} = 1.29 \frac{kg}{m^{3} }

\rho_{al} = 2.7× 10^{3} \frac{kg}{m^{3} }

The true mass of the aluminium is given by

m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }

Put all the values in above equation we get

m_{alu} = \frac{(2700)(4.5)}{2700-1.29}

m_{alu} = 4.5021 kg

Therefore the true weight of the aluminium is m_{alu} = 4.5021 kg

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2 years ago
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The third statement is correct.

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