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Julli [10]
3 years ago
8

When making maps of the large-scale universe, astronomers estimate distances to the vast majority of galaxies by using:

Physics
1 answer:
Vesnalui [34]3 years ago
4 0

Answer:

<em>The comoving distance and the proper distance scale</em>

<em></em>

Explanation:

The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster.  Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.

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1. The acronym AUDIENCE serves to remind you of what to consider when analyzing an audience. In the acronym, what does the C sta
Tom [10]
The correct answers are as follows:
1. D [CONTEXT]
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2. EXPECTATION. NEEDS.
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7 0
3 years ago
Read 2 more answers
A student plucks a fixed-end string, creating a standing wave with 6.00 nodes (including any nodes at the ends). The string is t
Vesna [10]

1) 2.5 wavelengths

2) 0.208 m

3) 1731 Hz

Explanation:

1)

Standing waves are waves that do not propagate, but instead the particles of the medium just oscillate around a fixed position. Examples of standing waves are the waves produced on a string with fixed ends.

The points of a standing wave in which the amplitude of the oscillation is always zero are called nodes.

The two fixed ends of the string are two nodes. In this problem, we have a total of 6 nodes along the string: this means that there are 4 additional nodes apart from the two ends of the string.

Therefore, this also means that the string oscillate in 5 different segments.

One wavelength is equal to 2 segments of the oscillation: therefore, since here there are 5 segments, this means that the number of wavelengths that we have in this string is

n=\frac{5}{2}=2.5

2)

The wavelength of a wave is the distance between two consecutive crests (or throughs) of the wave.

The wavelength of a standing wave can be also measured as the distance between the nth-node and the (n+2)-th node: so, basically, the wavelength in a standing wave is twice the distance between two nodes:

\lambda = 2 d

where

\lambda is the wavelength

d is the distance between two nodes

Here the length of the string is

L = 0.520 m

And since it oscillates in 5 segments, the  distance between two nodes is

d=\frac{L}{5}=\frac{0.520}{5}=0.104 m

And therefore, the wavelength is

\lambda=2d=2(0.104)=0.208 m

3)

The frequency of a wave is the number of complete oscillations of the wave per second.

The frequency of a wave is related to its speed and wavelength by the wave equation:

v=f\lambda

where

v is the speed

f is the frequency

\lambda is the wavelength

In this problem:

v = 360 m/s is the speed of the wave

\lambda=0.208 m is the wavelength

Therefore, the frequency is

f=\frac{v}{\lambda}=\frac{360}{0.208}=1731 Hz

3 0
3 years ago
A 0.900 kg block is attached to a spring with spring constant 14.5 N/m . While the block is sitting at rest, a student hits it w
olga55 [171]

Answer:

1. A=0.0847\ m

2. v=0.050\ m.s^{-1}

Explanation:

Given:

  • mass of block, m=0.9\ kg
  • spring constant, k=14.5\ N.m^{-1}
  • maximum velocity of block, v_m=34\ cm.s^{-1}

1.

<u>We know from the energy of oscillating spring:</u>

E=\frac{1}{2}.k.A^2= \frac{1}{2}.m.v_m^2

\frac{1}{2}\times 14.5\times A^2=\frac{1}{2}\times 0.9\times 0.34^2

A=0.0847\ m

2.

Now, given that:

instantaneous position, x=0.15\times A=0.0127\ m

we find angular speed,

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{14.5}{0.9} }

\omega=4.013\ rad.s^{-1}

So we have

v=x.\omega

v=0.0127\times 4.013

v=0.051\ m.s^{-1}

7 0
4 years ago
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