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3 years ago

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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What are three elements that are in the same group

cluponka [151]

Iodine, chloride, and bromide

3 0

2 years ago

Different between progressive wave and stationary wave

patriot [66]

1) Progressive waves are the ones by which all the wave disturbances are carried on further and propelled or transferred forward.

1) Stationary waves are the ones by which all the wave disturbances are not carried on further and propelled or transferred forward.

2) In Progressive waves the energies are equally and efficiently transferred along the travelling waves. Every particle are transferring some kind of energy to a next further particle on the same path, basically most of the energies are lost because of which there's no energy acquired by it.

2) In Stationary or standing waves there's no absolute transfer of any significant amount of energies which are not transferred along a path of the wave. Particles in stationery waves are giving and contributing in energy submission and also acquire some of the energy back because of which the net transferring of energies between the particles in a specific period as nullified.

3) Phases of the progressive waves of the particles in these waves are varying in a continuous manner and have changing values between them.

3) Phases of the Stationary or standing waves of the particles in these waves are not changing and always same to the contrary opposite when placed between the consecutively running sets of nodes (Between two nodes of particles).

4) Progressive waves have no particles which show they're having a rest phase or a permanent rest phase in a medium (particle medium).

4) Stationary waves have significant amount of particles of the medium to show that there having a rest phase or a permanent rest phase at the nodes of those particles.

5) Amplitudes of Progressive waves are totally and completely different particles are neutral and are having same values.

5) Amplitudes of Stationary or Standing waves of the particles in between those tow consecutively sets of nodes in between them and antinodes provided are having different values and vary much more progressively.

6) All of the particles in Progressive waves containing it don't specially cross their given mean positions in a simultaneous manner.

6) All of the particles in Stationary waves containing it frequently and steadily cross their given mean positions in a simultaneous manner.

7) In Progressive waves the particles don't show any attainment of a displacement provided in a maximum amount in a simulations manner.

7) In Stationary waves the particles are showing and exhibiting the attainments of various displacements in a maximum amount in a simultaneous manner.

8) Maximum velocities achieved by Progressive waves are indeed same or similar for all the given particles when they're showing a passing of those given mean positions.

8) Maximum velocities achieved by Stationary waves of those particles when they're crossing their given mean positions are in a continuity of increasement for those particles between those "nodes" and of course the consecutively set "antinodes" further which it's showing a significant decreasement after it corresponds and reaches the second or usually the next nearest node.

9) Progressive waves have crest and troughs in their waves that're moving into a forward direction.

9) Stationary waves have crest and troughs in their waves that're appearing and disappearing in same positions or regions that is , not moving forward in a same direction.

Read more on Brainly.in - https://brainly.in/question/1959503#readmore

7 0

3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced: