You observe a distant galaxy. you find that a spectral line normally found in the visible part of the spectrum is shifted toward the infrared. what do you conclude?

Answers

Answer 1

Answer: The galaxy is moving away from you.

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Jason hits a volleyball with an initial velocity of 6 meters per second straight up. if the back starts at 2 meters above the floor how long will the ball be in the air

An object traveling in a straight line accelerates. What will definitely happen due to the acceleration?

Answers

"Acceleration" means any change in speed or direction of motion.
If the object is known to be accelerating even though it's traveling
in a straight line, then we know that its speed must be changing.

In scientific terms, how is power related to work? A.Power is the ability to do work.

B.Power is the force of work over a certain distance.

C.Power is the rate at which work is accomplished.

D.Power is the energy supplied in a force.

Answers

C. the rate at which work is done. hope i helped!

Name three elements that are good conductors of electricity

Answers

Silver the very best conductorCopper the industrial choiceIron

A child is given an initial push on a rope swing. On the first swing, the rope swings through an arc of 12 feet. On each successive swing, the length of the arc is 80% of the previous length. After 14 swings, what total length will the rope have swung? When the child stops swinging, what total length will the rope have swung?

Answers

Answer:

After 14 swings, total length = 57.35 feet.

When the child stops swinging, Total length = 60 feet.

Explanation:

The total length of the rope after 14 swings form a geometric progression which is also known as exponential sequence.

The sum of the term in a Geometry progression is

Sₙ = a(1-rⁿ)/1-r.................(1)

Where Sₙ = sum of the nth term, a= first term, n= number of term, r= common ratio.

n=14, a= 12 feet, r=80% = 0.8.

Substituting the values above into equation(1)

S₁₄ = 12(1-0.8¹⁴)/1-0.8

S₁₄= 12(1-0.04398)/1-0.8

S₁₄= 12(0.95602)/0.2

S₁₄ = 11.47/0.2

S₁₄= 57.35 feet

∴ After 14 swings, the total length the rope will swing is = 57.35 feet.

The total length of the rope when the child stop swinging = sum to infinity of the Geometry progression( exponential sequence).

The sum to infinity of an exponential sequence

S = a/1-r

Where a= first term= 12 feet, r= common ratio = 0.8.

∴ S= 12/1-0.8

S= 12/0.2

S = 60 feet

When the boy stops swinging, the total length the rope have swung = 60 feet.

During the refrigeration cycle, the refrigerant absorbs and releases thermal energy by vaporizing and condensing.A.) True
B.) False

Answers

True.... the refrigerant is superheated from the end of the evaporator, through the compressor, up to the beginning of the condenser

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on the other to be LaTeX: 1.35\times10^{-4}N1.35 × 10 − 4 N when they are 20 cm apart. Accidentally, one the the experimenters causes the balls to collide and then repositions them 20 cm apart . Now the repulsive force is found to be LaTeX: 1.406\times10^{-4}N1.406 × 10 − 4 N. What are the initial charges on the two metal balls?

Answers

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=(1)/(4\pi\spsilon_0)(q_1q_2)/(d^2)-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$(1)/(4\pi\spsilon_0)=9*10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35*10^(-4)$ N

So, from equation (i)

$1.35*10^(-4)=9*10^9(q_1q_2)/((0.2)^2)$

$\Rightarrow q_1q_2=6*10^(-16)\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$(q_1+q_2)/(2)$

d=20cm=0.2m, and $F_c=1.406*10^(-4)$ N

So, from equation (i)

$1.406*10^(-4)=9*10^9(\left((q_1+q_2)/(2)\right)^2)/((0.2)^2)$

$\Rightarrow (q_1+q_2)^2=2.50*10^(-15)$

$\Rightarrow q_1+q_2=\pm5* 10^(-8)$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5* 10^(-8)\;\cdots(iii)$