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3 years ago

Why does the presence of redshift in spectra of galaxies support an understanding of an expanding universe?

Physics

1 answer:

xenn [34]3 years ago

7 0

Explanation:

Doppler's Effect: If a source is moving towards the observer the frequency of the waves emitted by the source will increase( wavelength will decrease). If the source is moving away from the observer the frequency of the waves emitted by the source will decrease( wavelength will increase). The best example of this is the changing frequency of siren of an ambulance as it moves towards us and then away from us.

The same phenomenon can be observed in celestial objects. When the object will move towards us the wavelength will decrease thus the spectrum will be shifted towards the blue part. This is known as blue shift.

If the object moves away from us the wavelength will increase and thus the spectrum will be shifted towards red part. This is know as Red-shift. This is because red has longer wavelength (620 - 750nm) than blue (450-495nm)

Thus if a galaxy shows us red-shifted spectra it means that it is moving away from us. we have observed this phenomenon in various galaxies thus supporting the expanding universe theory.

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A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of

Blizzard [7]

Answer:

the strength of the magnetic field is 3 x 10⁻⁵ T

Explanation:

Given;

velocity of the cosmic ray, v = 5 x 10⁷ m/s

force experienced by the ray, f = 1.7 x 10⁻¹⁶ N

angle between the ray's velocity and the magnetic field, θ = 45⁰

The strength of the magnetic field is calculated as;

$F = qvB \ sin(\theta)\\\\B = \frac{F}{qv\times sin(\theta)} \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 \times 10^{-19} \ C\\\\B = \frac{1.7\times 10^{-16}}{(1.602 \times 10^{-19})\times (5\times 10^7) \times sin \ (45)} \\\\B = 3 \times 10^{-5} \ T$

Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T

7 0

2 years ago

Which is the product of cellular respiration? A. ATTP B. light C. oxygen D.sugar

MrMuchimi

The answer to the question is A

7 0

2 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

LC Circuit

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

2 years ago

A person must pull a rope 20 meters with a force of 200 N what is the work input?

tatiyna

Work done = force * distance
work done = 200 * 20
work done = 4000J

4 0

2 years ago

What would be the weight of the moon if it were resting on the surface of the earth

kari74 [83]

We need to be careful here.
The calculation of the gravitational force between two objects
refers to the distance between their centers.
The minimum possible distance between the Earth's and moon's
centers is the sum of their radii (radiuses).

Earth's radius . . . . . 6,360 km = 6.36 x 10⁶ meters
Moon's radius . . . . . 1,738 km = 1.738 x 10⁶ meters
Sum of their radii = 8.098 x 10⁶ meters

Also:
Earth's mass . . . . . 5.972 x 10²⁴ kg
Moon's mass . . . . . 7.348 x 10²² kg

and now we're ready to go !

 Gravitational force =

 G M₁ M₂ / R²

= (6.67 x 10⁻¹¹ N-m²/kg²)(5.972 x 10²⁴ kg)(7.348 x 10²² kg)/(8.098 x 10⁶ m)²

= (6.67 · 5.972 · 7.348 / 8.098²) · (10²³) Newtons

= (I get ...) 4.463 x 10²³ Newtons

That's almost exactly 10²³ pounds

 = 50,153,000,000,000,000,000 tons.

Those are big numbers.
All I can say is: I wouldn't exactly call that "resting" on the surface".

7 0

2 years ago