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RideAnS [48]
3 years ago
11

I have been trying to do number 2 and I just don’t know what the answer is if you can help me thank you so mush

Physics
1 answer:
skelet666 [1.2K]3 years ago
4 0

Answer:

Strong Positive Linear Line Correlation

Explanation:

According to the diagram, there is a strong positive linear line correlation between velocity and distance of galaxies. Hubble showed that galaxies are receding away from people with a velocity that is proportional to their distance from people more distant galaxies recede faster than nearby galaxies.

About the diagram: Velocity is the distance relation among extra-galactic nebulae. Radial velocities is corrected for solar motion, are plotted against distances estimated from involved stars and mean luminosities of nebulae in a cluster. The black discs and full line represent the solution for solar motion by using the nebulae individually; the circles and broken line represent the solution combining the nebulae into groups; the cross represents the mean velocity corresponding to the mean distance of 22 nebulae whose distances could not be estimated individually.

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For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

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Answer:

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Explanation:

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Reviewing the correct answers is: Decrease the transparency of the atmosphere

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