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zaharov [31]
1 year ago
14

In 6–8 sentences, write a summary of what you learned from this activity. Include an explanation of what causes the spectral shi

fts you observed. Also, explain how scientists might use spectra analysis to support the big bang theory.
Physics
1 answer:
QveST [7]1 year ago
7 0

Answer :First part is what you learned so write what you learned then the other parts is The evidence that the universe is expanding comes with something called the red shift of light. Light travels to Earth from other galaxies. As the light from that galaxy gets closer to Earth, the distance between Earth and the galaxy increases, which causes the wavelength of that light to get longer.

Explanation:

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4. The value of the before and after-collision momentum of two colliding objects is shown in the
ExtremeBDS [4]

Answer:A) WHICH is 0kgm/s

Explanation:

3 0
3 years ago
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
MatroZZZ [7]

Answer:

The wavelength is \lambda_2 =  534 *10^{-9} \ m

Explanation:

From the question we are told that

   The wavelength of the first light is  \lambda _ 1 =  587 \ nm

    The order of the first light that is being considered is  m_1  =  10

     The order of the second light that is being considered is  m_2  =  11

Generally the distance between the fringes for the first light is mathematically represented as

      y_1 =  \frac{ m_1  * \lambda_1 *  D}{d}

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         y_2 =  \frac{ m_2  * \lambda_2 *  D}{d}

Now given that both of the light are passed through the same double slit

       \frac{y_1}{y_2}  =  \frac{\frac{m_1 *  \lambda_1 * D}{d}  }{\frac{m_2 *  \lambda_2 * D}{d}  } = 1

=>    \frac{ m_1 *  \lambda _1  }{ m_2  *  \lambda_2} =  1

=>     \lambda_2 =  \frac{m_1 *  \lambda_1}{m_2}

=>    \lambda_2 =  \frac{10  *   587 *10^{-9}}{11}

=>   \lambda_2 =  534 *10^{-9} \ m

4 0
3 years ago
(b) Two electrons in the same atom both have n = 3 and I = 1. Assume the
wolverine [178]

Answer:

We are social beings and we are going to be able to make a paper gun very very small it want to be a hacker in real life and we are going to be able to make

8 0
2 years ago
Which material is used to reduce friction ?
Iteru [2.4K]

Answer:MOST COMMON METHOD IS USING A LUBRICANT -

A lubricant is a substance, usually organic, introduced to reduce friction between surfaces in ... medical examination. It is mainly used to reduce friction and to contribute to a better and efficient functioning of a mechanism. ... For lubricant base oil use, the vegetable derived materials are preferred

OTHER METHODS-

There are a number of ways to reduce friction:

Make the surfaces smoother. ...

Lubrication is another way to make a surface smoother.

Make the object more streamlined.

Reduce the forces acting on the surfaces.

Reduce the contact between the surfaces.

PLEASE DO MARK ME AS THE BRAINLIEST :)

6 0
3 years ago
A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b
nekit [7.7K]
<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power  over one period. The instantaneous power can be found as:

p(t)=v(t)i(t)

So the average power is:

P=\frac{1}{T}\intop_{0}^{T}p(t)dt

But:

v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)

So:

P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}

P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}

In terms of RMS values:

V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI

7 0
3 years ago
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