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DerKrebs [107]
3 years ago
10

How was Edwin Hubble able to use his discovery of Cepheids in Andromeda to prove that the "spiral nebulae" were actually galaxie

s external to the Milky Way?
Physics
1 answer:
Ad libitum [116K]3 years ago
7 0

Answer:  The correct answer is :  From the period-luminosity relation for Cepheids, he was able to determine the distance to Andromeda and show that it was far outside the Milky Way Galaxy.

Explanation:  Hubble's law says that the recession velocity of a galaxy is directly proportional to its distance from us. Hubble measured the distance to the Andromeda galaxy by applying the period-luminosity relationship to Cepheid.

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Which federal agency protects ecosystems and supervisors public access the forests
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Answer:

United States forest service

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A child does 350J of work while pulling a box from the ground up to his tree house with a rope. The tree house is 5.2 m above th
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I think the answer to that questions is B.

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3 years ago
How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d
Archy [21]

Answer:

The voltage is   V =   0.993V_b

Explanation:

From the question we are told that

   The time that has passed is  t = \frac{\tau}{2}

 Here \tau is know as the time constant

    The voltage of the  power source is   V_b

Generally the voltage equation for charging a capacitor is mathematically represented as

       V =  V_b  [1 - e^{- \frac{t}{\tau} }]

=>   V =  V_b  [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]

=>   V =  V_b  [1 - e^{- \frac{\tau}{2\tau} }]

=>   V =  V_b  [1 - e^{- \frac{1}{2} }]

=>   V =   0.993V_b    

5 0
3 years ago
A 10.00 kg block is placed at the top of a long frictionless inclined plane angled at 37.9 degrees relative to the horizontal. T
NeX [460]

Answer

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Explanation:

4 0
2 years ago
The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm
Nesterboy [21]

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

8 0
3 years ago
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