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Rina8888 [55]
2 years ago
15

If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc

Physics
1 answer:
prohojiy [21]2 years ago
4 0

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

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hodyreva [135]

Answer:

a) 1.517\times10^{-11} s

b) 3.41 mm

Explanation:

a)

We take the speed of light, c = 3.0\times10^8 m/s and the refractive index of glass as 1.517.

Speed = distance/time

Time = distance/speed

Refractive index, n = speed of light in vacuum / speed of light in medium

n=\dfrac{c}{s}

s=\dfrac{c}{n}

t=\dfrac{d}{c/n}

t=\dfrac{dn}{c}

t=\dfrac{3\times10^{-3}\times1.517}{3.0\times10^8}

t=1.517\times10^{-11}

b)

We take the refractive index of water as 1.333.

Speed in water = speed in vacuum / refractive index of water

Distance = speed * time

d=s\times t

d=\dfrac{c}{n_w}\times \dfrac{3\times10^{-3}\times1.517}{c}

d=\dfrac{3\times10^{-3}\times 1.517}{1.333}

d = 3.41 mm

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3 years ago
9- Under what circumstances would a vector have components that are equal in
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Explanation:

c. if the vector is oriented at 0° from the X -axis.

6 0
2 years ago
If a 100-N net force acts on a 50-kg car, what will the acceleration of the car be?
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Divide each side by (mass):     

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Answer:

a. the core will spin faster.

Explanation:

By law of conservation of angular momentum

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v= speed of star

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f= final

since, size(R) of the star is reduced by factor of 10,000 and mass remains the same, the velocity must increase by the same factor to keep the angular momentum conserved.

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Some miners wish to remove water from a mine shaft. A pipe is lowered to the water 90 m below, and a negative pressure is applie
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Answer:

Explanation:

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