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Alona [7]
3 years ago
5

Redshift means stars and galaxies are emitting a color that is shifting toward the red end of the color spectrum. this indicates

the light's wavelengths are _____ and less frequent, and the star or galaxy is____?
Physics
2 answers:
iren [92.7K]3 years ago
7 0
Increasing ; moving away

This indicates the light's wavelengths are (increasing) and the star or galaxy is (moving away.)
Olin [163]3 years ago
6 0

Answer:

Redshift means stars and galaxies are emitting a color that is shifting toward the red end of the color spectrum. this indicates the light's wavelengths are <u>MORE</u> and less frequent, and the star or galaxy is <u>MOVING AWAY</u>

Explanation:

As we know that by Doppeler's effect of light when source is moving away from the observer then the frequency observed by the observer is given as

\frac{\Delta \nu}{\nu} = \frac{v}{c}

so here on moving away the frequency observed by observer will decrease and hence we can say that the wavelength of light must be increased

So here on increasing the observed wavelength we say it to be shifted towards higher side which means it is red shift

so here correct answer would be

Redshift means stars and galaxies are emitting a color that is shifting toward the red end of the color spectrum. this indicates the light's wavelengths are <u>MORE</u> and less frequent, and the star or galaxy is <u>MOVING AWAY</u>

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The force acting between the particles is

F=k \frac{Q_{1}Q_{2}}{r^2}
Then
Q_{2}= \frac{5.2 \times 10^-^5 \times 0.024^2}{ 9.0 \times 10^9 7.2 \times 10^-^8} =4.622 \times 10^-^1^1C




7 0
3 years ago
Read 2 more answers
An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
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Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

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3 years ago
in which kind of radioactive decay would the number of protons in the resulting nucleus be more than in the initial nucleus?
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Answer:

A related type of beta decay

Explanation:

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2 years ago
A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
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Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J              

Explanation:

We have given the player turntable initially rotating at speed of 33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec

Now speed is reduced by 75 %

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Time t = 5.5 sec

From first equation of motion we know that '

\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2

(a) Now final velocity \omega =0rad/sec

So time t to come in rest  t=\frac{0-3.49}{-0.158}=22sec

(b) The work done in coming rest is given by

\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J

4 0
3 years ago
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12345 [234]

Answer:

Explanation:

<u>Given:</u>

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The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by \epsilon_0

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So the the charge at the centre is calculated.

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3 years ago
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