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o-na [289]
3 years ago
8

the value of acceleration due to gravity option A is same on equator and poles option B is list on poles option C is list on equ

ator option d increase from pole to equator which is the answer​
Physics
1 answer:
stiks02 [169]3 years ago
8 0

Answer:

D. Increases from pole to equator

Explanation:

I majored in Science

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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.

6 0
3 years ago
Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.
Gelneren [198K]

Answer:

2.99\cdot 10^{11}C

Explanation:

The mass of one electron is

m_e = 9.11\cdot 10^{-31}kg

So the number of electrons contained in M=1.7 kg of mass is

N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}

The charge of one electron is

e=1.60\cdot 10^{-19} C

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C

8 0
3 years ago
What is the relation between acceleration due to gravity and radius of the earth? ​
emmasim [6.3K]

Answer:

As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.

Formula: g = GM/r2

Dimensional Formula: M0L1T-2

Values of g in SI: 9.806 ms-2

Explanation:

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8 0
2 years ago
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The diagram shows parts of a wave. A series of waves with an arrow passing through their centers. The highest point of one wave
Pachacha [2.7K]

Answer:

im not sure but i think its c

Explanation:

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3 years ago
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Scientific notation of 86400
Mandarinka [93]
8.64×10^4

this is 86400 in scientific notation

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3 years ago
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