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AnnyKZ [126]
3 years ago
11

We can detect 21-cm emission from other galaxies as well as from our own Galaxy. However, 21-cm emission from our own Galaxy fil

ls most of the sky, so we usually see both at once. How can we distinguish the extragalactic 21-cm emission from that arising in our own Galaxy? (Hint: Other galaxies are generally moving relative to the Milky Way.)
Physics
1 answer:
amid [387]3 years ago
6 0

Answer:

Explanation:

Galaxies are in constant motion with respect to each other . For example Andromeda galaxy is approaching our galaxy ( milky way ) at about 110 km /s . So we will observe blue shift in the spectrum of radiation coming from  this galaxy . In this way, we can distinguish between radiation coming from our galaxy and that coming from other galaxy . Spectrum of radiation coming from other galaxy must  have either red or blue  shift .

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I need help ! if anyone answers this its worth 40 points please help !!
Sliva [168]
Double displacement...I think
3 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
WILL GIVE BRAINLIEST
krek1111 [17]

Answer:

I think d

Explanation:

6 0
3 years ago
A bowling ball weighing 71.2 N is attached to the ceiling by a 3.30 m rope. The ball is pulled to one side and released; it then
wolverine [178]

Answer:

a. The acceleration of the bowling ball is 9.5 m/s² toward the center

b.  The tension in the rope is 140.24 N

Explanation:

given information:

ball weight,  W = 71.2 N

the length of rope, R = 3.30 m

ball speed, v = 5.60 m/s

a. The acceleration of the bowling ball, α

α = \frac{v^{2} }{R}

where

α = the acceleration

v = the speed

R = radius

thus

α = \frac{v^{2} }{R}

  = \frac{5.60^{2} }{3.3}

  = 9.5 m/s² toward the center

b. The tension in the rope?

according to the Newton's second law

ΣF = m a

where

F = force

m = mass

a = acceleration

so,

ΣF = m a

T - W = m a

T = m a + W

  = (W a/g) + W

  = (71.2 x 9.5/9.8) + 71.2

  = 140.24 N

4 0
3 years ago
A tow truck pulls a 2,000-kg car with a force of 4,000 N. what is the acceleration of the car?
ad-work [718]

Answer:

B

Explanation:

using f=ma

a=f/m

=4000/2000

=2 m/s2

8 0
3 years ago
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