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pshichka [43]
3 years ago
13

Explain why stellar parallax cannot be used to measure the distance to other galaxies.

Physics
1 answer:
Phoenix [80]3 years ago
8 0

Answer:

1. a) Astronomers use the parallax method to measure the distance to nearby stars, but

we can’t use it to measure the distance to stars in other galaxies. Why not? Why isn’t the

parallax method useful for measuring the distances to stars in other galaxies?

They are so distant that the parallax is too small to be measured since parallax varies

inversely with distance.

b) Instead of the parallax method, we use the standard candle method to measure the

distance to stars in other galaxies. In particular, we use the standard candle method to

measure the distances to Cepheid variable stars in other galaxies. What is special about

Cepheid variable stars that makes them useful for this purpose?

We can figure out their luminosities from their periods of variation. Then if we measure

their fluxes we can calculate their distances.

2. a) From what were the protons and electrons in your body made, and roughly when

were they made?

They were made from energy (or gamma rays) very soon after the big bang (in the first

second). 400,000 years later they got together to make hydrogen atoms.

b) From what were the carbon atoms in your body made, and where were they made?

They were not made in the big bang. They were made much later inside of stars or in

supernovae. They were made by fusion from lighter atoms.

3. Make two sketches of the Milky Way Galaxy, one an edge-on view and one a face-on

view, labeling the various parts of the galaxy.

You should have labeled the

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The teardrop could be an example as it was designed for that purpose, and most notably planes and such aero traveling vehicles
4 0
3 years ago
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Perfect symmetrical cone type volcanoes would have what type of drainage pattern
Charra [1.4K]

Perfect symmetrical cone type volcanoes would have radial drainage pattern.

A stratovolcano, also known as a composite volcano, is a conical volcano built up by many layers (strata) of hardened lava, tephra, pumice, and volcanic ash.

If you have any further questions, please don’t hesitate to ask again.

6 0
3 years ago
3) A dock worker pushes a 72 kg crate up a 2.0 m high,
Vlad [161]

Work done on the crate is 1411.2 J

Explanation:

Work done is defined as the product of force and the distance moved by the object. The unit of work done is in joules and denoted by the symbol J.

                                     Work done = F * d

where F represents the force and d represents the distance moved by the object.

mass = 72 kg , distance moved by the object is given by 2.0 m

Force F = mass * gravity = 72 * 9.8

             = 705.6 N =706 N.

Work done = 706 * 2.0 = 1412 J.

                   

7 0
3 years ago
4. How much work is done when you lift a 25 N box up to a 2 m shelf?
Semmy [17]

Answer:

<h2>50 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

force = 25 N

distance = 2 m

We have

workdone = 25 × 2 = 50

We have the final answer as

<h3>50 J</h3>

Hope this helps you

6 0
2 years ago
R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d
padilas [110]

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

brainly.com/question/12647190

#SPJ10

3 0
2 years ago
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