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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

You might be interested in

Rays traveling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus?

abruzzese [7]

Answer:

True

Explanation:

When a ray travelling parallel to the principle axis of a concave mirror then the light ray reflect out through the mirrors and passing through the focus.

When a light ray travelling through focus of a concave mirror then after reflection the light ray reflect out through the mirror and go parallel to principle axis.

Therefore, rays travelling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus.

It is true.

7 0

3 years ago

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How do scientists study pollen grains to help them understand climate change?

tino4ka555 [31]

Answer: A. by analyzing and making inferences about them

Explanation:

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3 years ago

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There is a refrigerator running in a room, the heat flowing into the refrigerator from the outside is 40 J/s, and the refrigerat

Mamont248 [21]

Answer:

(A) 140 j/sec (b) 1.26 K

Explanation:

We have given the heat heat flowing into the refrigerator = 40 J/sec

Work done = 40 W

(a) So the heat discharged from the refrigerator $=heat\ flowing\ in\ refrigerator+work\ done=40+100=140j/sec$

(b) Total heat absorbed =140 j/sec $=140\times 3600=504000j/hour$

Let the temperature be $\Delta T$

Heat absorbed per hour =504000 $[tex]=400\times 10^3\times \Delta T$

So $\Delta T=\frac{504000}{400000}=1.26K$

8 0

3 years ago

When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the

KonstantinChe [14]

Answer:

The ratio is KE : TM = 0.75

Explanation:

from the question we are told that

The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical energy of the mass is mathematically represented as

$TM = \frac{1}{2} * k * A^2$

Here k is the spring constant , A is the total displacement of the the mass from maximum compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

$TM = KE + PE$

=> $KE = TM - PE$

Here the potential energy of the mass is mathematically represented as

$PE = \frac{1}{ 2} * k * [ x ]^2$

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is

$x = \frac{A}{2}$

$PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2$

$KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2$

=> $KE = \frac{1}{2} * k * A^2 - \frac{1}{8} * k * A ^2$

=> $KE = 0.375 * k * A^2$

So the ratio of $KE : TM$ is mathematically represented as

$\frac{KE}{TM} = \frac{0.375 k A^2 }{0.5 k A^2}$

=> $\frac{KE}{TM} = 0.75$

3 0

3 years ago

A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

3 years ago