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valina [46]
3 years ago
5

consider the hypothetical observation irregular galaxies outside the local group are moving toward us. from part a, this observa

tion would contradict the idea of an expanding universe. why?
Physics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

spiral galaxies move away from us 10% faster than elliptical galaxies at the same distances;  irregular galaxies outside the Local Group are moving toward us;  galaxy speeds are faster in summer than in winter

Explanation:

This may help I'm not sure.

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A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy
SashulF [63]

Answer:

5.444\times 10^{-4}

Explanation:

The momentum of the neutron before and after the decay  is the same since there's no external force.

P_{sys}=const\\\\P=mv\\\\K=0.5mv^2

#The neutron is initially at rest, so after the decay:

P_A+P_B=0\\\\P_A=-P_B

#After decay, the proton has +ve direction  with a velocity v_Awhile the electron moves in a negative direction with a velocity v_B

Therefore:

P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B

Let the energy released during the decay be Q:

Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}

Hence,Kp/Ktot is 5.444x10^(-4)

4 1
3 years ago
The annoying sound from a mosquito is produced when it beats its wings at the average rate of 600. wingbeats per second. What is
jok3333 [9.3K]
The frequency is exactly the rate at which the bug beats its wings.
If that "600" that you mentioned is 600 beats per second, then
THAT's the frequency . . . 600 per second. (600 Hz) 
3 0
4 years ago
A thin film of cooking oil ( n = 1.47 ) is spread on a puddle of water ( n = 1.35 ) . What is the minimum thickness D min of the
zlopas [31]

Answer:

The minimum thickness of the oil is 77.55 nm

Explanation:

Given:

Refractive index of oil n_{o} = 1.47

Refractive index of water n_{w} = 1.35

Wavelength of light \lambda= 456 \times 10^{-9} m

From the equation of thin film interference,

The minimum thickness is given by,

    2n_{o} t = (n+\frac{1}{2}) \lambda

Where n = 0,1,2,3.........,t = thickness

Here we have to find minimum thickness so we use n = 0

     2n_{o} t  =( 0+\frac{1}{2} )\lambda

   t = \frac{\lambda }{4 n_{o} }

   t = \frac{456 \times 10^{-9} }{4 \times 1.47}

   t = 77.55 \times 10^{-9} m

   t  = 77.55 nm

Therefore, the minimum thickness of the oil is 77.55 nm

7 0
4 years ago
After creating a question and forming a hypothesis, what is the next step in the scientific method
Keith_Richards [23]
The next step is to create research
3 0
3 years ago
5 litres of alcohol have a mass of 4kg. calculate the density of alcohol in g/cm.​
Aleks04 [339]

Answer: 0.8 g/cm

Explanation:

p= m/V

= 4 kg/ 5 liter

= 0.8

3 0
3 years ago
Read 2 more answers
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