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melamori03 [73]
2 years ago
5

Suppose that you measure a galaxy's redshift, and from the redshift you determine that its recession velocity is 30,000 (3×10^4)

kilometers per second. According to Hubble's law, approximately how far away is the galaxy?
Physics
1 answer:
anyanavicka [17]2 years ago
5 0

Answer:

1.4 billion light years away

Explanation:

v = Recessional velocity = 30000 km/s[/tex]

H_0 = Hubble constant = \frac{65}{3.2\times 10^6}\ ly

D = Distance to the galaxy

According to Hubble's law

v=H_0D\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{3\times 10^{4}}{65}\times 3.2\times 10^6\\\Rightarrow D=1476923076.92307\ ly\\\Rightarrow D=1.4\times 10^9\ ly

The galaxy is 1.4 billion light years away

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<span>Vf = Vi + at
</span>Vf = 0 + (-9.8 m/s^2) (7 s)
Vf = -68.6 m/s

The negative sign denotes the direction of the ball.
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3 years ago
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Density=2g/mL and volume=20mL what is mass
leva [86]

Answer:

40g

Explanation:

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How does radiation differ from conduction?
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b

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They create a heat engine where the hot reservoir is filled with water and steam at equilibrium, and the cold reservoir is fille
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The efficiency of Carnot's heat engine is 26.8 %.

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temperature of cold reservoir, Tc = 0 degree C = 273 K

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4 0
3 years ago
An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
Katen [24]
<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

Where:

y  is the instrument's final position  

y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

t is the time the parabolic movement lasts

g=2.5\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of planet X.

As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

Finally:

t=12s

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3 years ago
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