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77julia77 [94]
3 years ago
12

The entropy of a hot baked potato decreases as it cools. is this a violation of the increase of entropy principle? explain.

Physics
1 answer:
marshall27 [118]3 years ago
6 0
<span>This is not a violation of entropy principle, because this law is not violated in any way as it applies to both real an ideal situations independently. The entropy of a real system always raises and is an irreversible process.</span>
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A guitar string is 90 cm long and has a mass of 3.7 g . The distance from the bridge to the support post is L=62cm, and the stri
zvonat [6]

To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

Here

n = Number of node

T = Tension

\mu = Linear density

L = Length

Replacing the values in the frequency and value of n is one for fundamental overtone

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})

\mathbf{f = 281.2Hz}

Similarly plug in 2 for n for first overtone and determine the value of frequency

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})

\mathbf{f = 562.4Hz}

Similarly plug in 3 for n for first overtone and determine the value of frequency

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)

\mathbf{f= 843.7Hz}

4 0
3 years ago
What is 18 degrees celsius in fahrenheit ?
Butoxors [25]
18 degree is equal to 64.4 fahrenheit

4 0
3 years ago
Does this equation satisfy the Laws of Conservation of Matter and Mass?
Monica [59]

Answer:

yes

Explanation:

it does

5 0
3 years ago
All the organelles in a cell work together to help the cell and the whole organism function. Some diseases are caused by the fai
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Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black
oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

Q = FA\sigma\Delta T^4

Where,

F =View Factor

A = Cross sectional Area

\sigma = Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K

The view factor between two coaxial parallel disks would be

\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33

\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75

Then the view factor between base to top surface of the cylinder becomes F_{12} = 0.26. From the summation rule

F_{13} = 1-0.26

F_{13} = 0.74

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}

\dot{Q_3} = 2\dot{Q_{13}}

\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)

\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)

\dot{Q_3} = 780.76W

Therefore the rate heat radiation is 780.76W

5 0
3 years ago
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