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Nonamiya [84]
2 years ago
7

8. What is the frequency of the standing wave shown?

Physics
1 answer:
densk [106]2 years ago
3 0

Answer:

while using brainly app. there is an option of uploading image of diagram, graph or plot. plz use that feature to upload image of the standing wave so that your question can be properly answeres

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A thin-walled cylindrical steel water storage tank 30 ft in diameter and 62 ft long is oriented with its longitudinal axis verti
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ρ

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55.0

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Explanation:

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3 years ago
Can you guys please help me out and it would me great if you could explain it too!!
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When sunlight shines on a leaf the leaf looks green why does the leaf look green
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The correct answer is the third, It reflects the green light waves and absorbs most of the rest.
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3 years ago
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Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin
Vedmedyk [2.9K]

Answer:

The correct answer is 2.8*10^{-5}ms^{-1}

Explanation:

The formula for the electron drift speed is given as follows,

u=I/nAq

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is 8.93/63.5 molcm^{-3}. Converting this number to m³ using very elementary unit conversion we get 140384molm^{-3}. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}

if we convert the area from mm³ to m³ we get A=80*10^{-6}m^{2}.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}

which is our final answer.

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3 years ago
15. What is a thermograph?
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