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nekit [7.7K]
3 years ago
7

What is a consequence of the second law of thermodynamics?

Physics
1 answer:
marshall27 [118]3 years ago
4 0

Answer:Increase in the entropy of the universe are the result of chemical reactions

Explanation:

The second law tells about the quality of energy such that any isolated system tends to become more disordered. Any natural Process occurring on its own is irreversible and entropy is increasing for that process.  

The entropy of the Universe is increasing which is the result of chemical reactions. Any chemical reaction which increases the number of gas molecules will generate more entropy. Entropy in simple meaning is the order of randomness.

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Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f
gulaghasi [49]

Answer:

A)3.8196 * 10^{9} Newtons        B) 2.153 * 10^{9} Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as  120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = πR^{2}.

Area Expansivity β =    \frac{Change in Area}{Original Area * Temperature Rise}

Area Expansivity β = 2α

α =  16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = \frac{0.2m}{2} = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = \frac{0.15}{2} = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * 0.1^{2} =  0.0314m^{2}

Area of Rod BC = π * 0.075^{2} = 0.0177m^{2}.

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9*10^{-6} * 0.0314 * 30 = 3.183 * 10^{-5}m^{2}.

For Rod BC we have = 2 * 16.9*10^{-6} * 0.0177 * 30 = 1.794∈-5m^{2}.

The forces on each rods will be given by.

Force AB = Pressure * Area  of Rod AB

                 = 120GPa * 3.183 * 10^{-5} gives 3.8196 * 10^{9} Newtons.

Force BC = Pressure * Area of Rod BC

                = 120GPa * 1.794 * 10^{-5} gives 2.153 * 10^{9} Newtons

7 0
3 years ago
Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a
Nat2105 [25]

Answer:

Explanation:

Given

diameter d=20 \mu m

density \rho =1300 kg/m^3

frequency \nu =150 Hz

Length of silk strand L=14 cm

Velocity in the string is as follows

\nu =\sqrt{\frac{T}{\mu }}

The expression for Fundamental Frequency

f=\frac{\nu }{2l}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\frac{m}{l}}}

f=\frac{1}{2l}\times \sqrt{\frac{Tl}{\rho V}}

Squaring

f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}

T=4\rho \cdot A\cdot l^2\cdot f^2

T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2

T=7.2\times 10^{-4} N

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Answer:

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Two pieces of evidence for plate tectonics?
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