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Lena [83]
3 years ago
14

Which one of the following is NOT an evidence of the Big Bang Theory? A. most objects in space are moving away from one another

B. cosmic microwave background radiation C. the light from objects in space is “blue shifted” D. most of the universe is made from light elements, hydrogen and helium
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0
Hello there!

The correct answer to your question is c.

This is because one of the main claims to the Big Bang Theory is that "the universe is constantly expanding outwards."

When various objects, say galaxies, recede from us, their light's wavelength is elongated. This, in turn, results as the galaxy's light to look red.

When these objects are moving <em>closer</em> to us, it shortens the wavelength, creating a blue light. In terms of the Big Bang Theory, this should not happen, because everything is moving <u>out</u>.

I hope this helps!
Brady
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Two microwave signals of nearly equal wavelengths can generate a beat frequency if both are directed onto the same microwave det
alekssr [168]

Answer:

1.5106 cm

Explanation:

The beat frequency is equal to the absolute value of the difference between the frequencies of the two signals:

f_B = |f_1 - f_2|

using the wave equation, we can re-write each frequency as

f=\frac{c}{\lambda}

where c is the speed of light and \lambda is the wavelength. Therefore,

f_B = |\frac{c}{\lambda_1}-\frac{c}{\lambda_2}|

where:

f_B = 140 MHz = 140\cdot 10^6 Hz is the beat frequency

\lambda_1 = 1.50 cm = 0.015 m is the wavelength of the first generator

\lambda_2 is the wavelength of the second generator

We also know that the second generator emits the longer wavelength, so we already know that the term inside the module is positive. Therefore, we can now solve for \lambda_2:

f_B = c(\frac{1}{\lambda_1}-\frac{1}{\lambda_2})\\\lambda_2=(\frac{1}{\lambda_1}-\frac{f_B}{c})^{-1}=(\frac{1}{0.015}-\frac{140\cdot 10^6}{3\cdot 10^8})^{-1}=0.015106 m = 1.5106 cm

4 0
3 years ago
Use the accompanying seismogram to answer which of the three types of seismic waves reached the seismograph first.
UkoKoshka [18]

Answer:

Primary waves (P-waves)

Explanation:

Due to excess of the energy inside the earth when the tectonic plates begin to slide or fracture then the energy is released in the form of seismic waves, this causes the earthquake.

<u>Two types of seismic waves are generally responsible for the earth quakes:</u>

  1. body waves
  2. surface waves

Body waves are of two types:

Primary waves (P-waves)

These are the fastest of all the waves involved in the earth-quake which travel at a speed of 1.6 km to 8 km per second.

They can pass trough solids, liquids and gases. They arrive at the surface as an instant thud.

Secondary waves (S-waves)

They can only pass through the solids and they move slower than the P-waves.

As S-waves move, they displace the rock particles, pushing them outwards perpendicular to the wave-path that leads to the earthquake-related first rolling period.

Surface waves (L-waves/ long waves)

  • These waves move along the surface of the earth. They are responsible for the earthquake's carnage.
  • They move up and down the Earth's surface, rocking the foundations of man-made structures.
  • Surface waves are slowest of the three waves, which means that they are the last to arrive. So at the end of an earthquake usually comes the most powerful shaking.
6 0
3 years ago
The intensity at a distance of 6 m from a source that is radiating equally in all directions is 6.0\times 10^{-10} W/m^2. What i
Sophie [7]

Answer:

P = 271 nW

Explanation:

  • If the source is radiating equally in all directions, it can be treated as a point source, so all points located at the same distance of the source, have the same intensity I, which is related to the power by the following expression:

       I = \frac{P}{A} =\frac{P}{4*\pi *r^{2} } (1)

  • Solving for P, we get:

       P = I*4*\pi *r^{2} = 6.0e-10 W/m2 * 4 * \pi *(6m)^{2} =271 nW (2)

3 0
3 years ago
ANSWER SOOONN PLZ!!
Ket [755]

c it is both a and b

3 0
3 years ago
What is the net work doneon the object over the distance shown?
GuDViN [60]

A)F_0d

Explanation

If you graph the force on an object as a function of the position of that object, then the area under the curve will equal the work done on that object, so we need to find the area under the function to find the work

Step 1

find the area under the function.

so

Area:

\text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red}\begin{gathered} \text{the area of a rectangle is given by} \\ A_{rec}=lenght\cdot widht \\ \text{and} \\ \text{the area of a triangle is given by:} \\ A_{tr}=\frac{base\cdot height}{2} \end{gathered}

so

\begin{gathered} \text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red} \\ \text{replace} \\ \text{Area}=(F_0\cdot d)+\frac{(F_0\cdot d)}{2}-\frac{(F_0\cdot d)}{2} \\ \text{Area}=(F_0\cdot d) \\ Area=F_0d \end{gathered}

therefore, the answer is

A)F_0d

I hope this helps you

4 0
1 year ago
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