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ioda
3 years ago
5

What causes thermal expnsion?

Physics
2 answers:
VLD [36.1K]3 years ago
4 0

Answer:

Thermal expansion occurs when an object expands and becomes larger due to a change in that specific object's temperature. Temperature is the average kinetic (basically movement) energy of the molecules in a substance. A higher temperature means that the molecules are moving faster on average.

Dmitry_Shevchenko [17]3 years ago
4 0

Answer: Thermal expansion occurs when an object expands and becomes larger.

Explanation: This occurs when a change ( increase or decrease )

Forms in an objects temperature

For example, Increase in size of a material due to added heat.

Decrease in size of material when heat is removed.

You might be interested in
Match the lithification processes.
butalik [34]

Explanation:

Recrystallization: contact pressure causing grains to "fuse" together

Cementation : precipitation of bonding agents between grains

Compaction : increase in density due to weight of overburden

Lithification is the process by which sediments are converted into sedimentary rocks. During this process, recrystallication, compaction and cementation of mineral grains occur.

The process starts with the compaction of sediments. The over burden weight of new sediments in the basin adds to the one originally deposited. This compresses the sediment. The volume of reduced and the density increases.

Recrystallization follows suit as the contact pressure of grains makes them fuse together. It is more like reworking of sediments. In this process, cementing materials can precipitate and cause sediments to be more fused together.

This is why most sediment are made up of clasts in a matrix of cementing materials.

learn more:

sedimentary rocks brainly.com/question/9131992

#learnwithBrainly

4 0
3 years ago
If a truck runs into a wall and stops, the truck loses momentum. Because
nignag [31]
The momentum goes to the wall
6 0
3 years ago
Examine figure 26-1 which area is the convection zone<br><br>A. A<br>B. B<br>C. C<br>D. D
kykrilka [37]
I think the answer is B
4 0
4 years ago
Read 2 more answers
Two airplanes leave an airport at the same time.The velocity of the first airplane is 700 m/h at a heading of 31.3 the velocity
MArishka [77]

Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity

v_{12} = v_1 - v_2

here

speed of first plane is 700 mi/h at 31.3 degree

v_1 = 700 cos31.3\hat i + 700 sin31.3\hat j

v_1 = 598.12\hat i + 363.7\hat j

speed of second plane is 570 mi/h at 134 degree

v_2 = 570 cos134 \hat i + 570 sin134 \hat j

v_2 = -396\hat i + 410\hat j

now the relative velocity is given as

v_{12} = (598.12 + 396)\hat i + (363.7 - 410)\hat j

v_{12} =994.12\hat i -46.3 \hat j

now the distance between them is given as

d = v* t

d = (994.12 \hat i - 46.3 \hat j)* 3

d = 2982.36\hat i - 138.9\hat j

so the magnitude of the distance is given as

d = \sqrt{2982.36^2 + 138.9^2}

d = 2985.6 miles

so the distance between them is 2985.6 miles

5 0
3 years ago
Read 2 more answers
calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame
kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2

0 - p × Area for trachea =

\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

⇒-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}

           \frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0
3 years ago
Read 2 more answers
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