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densk [106]
3 years ago
13

If a galaxy has an apparent radial velocity of 2000 km/s and the Hubble constant is 70 km/s/Mpc, how far away is the galaxy

Physics
1 answer:
ddd [48]3 years ago
3 0

Answer:

28.57 Mpc

Explanation:

This question is going to be solved by applying Hubble's Law.

This Hubble's Law is actually an observation in physical cosmology. This observation makes it clear that galaxies are moving away from the Earth, and are doing so at speeds proportional to their distance. This essentially means that the farther they are from the Earth, the faster they are moving away from Earth.

It is represented by this formula

v = H(0)D, where

v = speed

H(0) = Constant of proportionality, or otherwise, Hubble's constant.

D = Distance to a galaxy

Applying the given parameters to the formula, we have

v = H(0).D

D = v / H(0)

D = 2000 / 70

D = 28.57 Mpc

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An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?
ICE Princess25 [194]

<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>

<span>y = 0 + 0t + ½gt² </span>
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<span>t = √(2y/g) </span>

<span>in the ft - lb - s system </span>

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7 0
4 years ago
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Imagine you have just witnessed a small avalanche on a mountain while skiing, and two slushy snowballs just crashed together in
inn [45]
We have that the momentum p is given by the formula p=mv where m is the mass and v is the velocity. Since for A p=-14kgm/s and m=7, we have that the velocity is -14/7=-2m/s. Hence its speed is 2 m/s.
For b we have that p=15kgm/s and v=3m/s. Because m=p/v, we have m=3kg.
We also have that the momentum is conserved in this system. Hence, the net sum of the momentum of the 2 snowballs equals the momentum of the single giant ball. Hence, p(total)=p(combined)=-14+15=1kgm/s (momentum is a vector; the positive sign means that it tends to the positive direction).

6 0
3 years ago
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Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i
Rashid [163]

Answer:

<em>a) 0.72 V</em>

<em>b) 19.2 mA</em>

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = N_{p} = 500 turns

input voltage = V_{p} = 120 V

number of secondary turns = N_{s} = 3 turns

output voltage = V_{s} = ?

using the equation for a transformer

\frac{V_{s} }{V_{p} }  = \frac{N_{s} }{N_{p} }

substituting values, we have

\frac{V_{s} }{120 }  = \frac{3 }{500} }

500V_{p}  = 120*3\\500V_{p} = 360

V_{p} = 360/500 =<em> 0.72 V</em>

<em></em>

b) by law of energy conservation,

I_{P}V_{p} = I_{s}V_{s}

where

I_{p} = input current = ?

I_{s} = output voltage = 3.2 A

V_{s} = output voltage = 0.72 V

V_{p} = input voltage = 120 V

substituting values, we have

120I_{p} = 3.2 x 0.72

120I_{p} = 2.304

I_{p}  = 2.304/120 = 0.0192 A

= <em>19.2 mA</em>

<em></em>

c) power input = I_{p} V_{p}

==> 0.0192 x 120 = <em>2.304 Watts</em>

7 0
3 years ago
Use Eq. cosϕ=R/Z to show that the average power delivered by the source in an L−R−C series circuit is given by Pav = I^2rmsR .
Evgen [1.6K]

Answer:

Explanation:

In a L C R circuit, the average power is given by

P_{av}=V_{rms}I_{rms}Cos\phi

As given in the question

CosФ = R / Z

And we know that

V_{rms}=I_{rms}\times Z

So

P_{av}=I_{rms}\times Z\times I_{rms}\times \frac{R}{Z}

P_{av}=I_{rms}^{2}\times R

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xxMikexx [17]

Answer:

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Explanation:

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