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2 years ago

A galaxy with a lot of dark matter would have a high mass-to-light ratio compared to the Sun. True False

Physics

1 answer:

Ivenika [448]2 years ago

8 0

Yes, a galaxy with a lot of dark matter would have a high mass-to-light ratio compared to the Sun.

<h3>Contribution of dark matter in mass to light ratio?</h3>

This largely invisible matter such as dark matter adds to the mass of the galaxy while it has no contribution in its luminosity which increases the mass-to-light ratio. If dark invisible matter is present in a galaxy, its mass-to-light ratio can be as high.

So we can conclude that the statement is true.

Learn more about galaxy here: brainly.com/question/13956361

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A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

4 years ago

What is the route number for ( 697 and 129 )

kramer

540.3 cause i divided them

8 0

3 years ago

To determine the pressure in a fluid at a given depth with the air-filled cartesian diver, we can employ Boyle's law, which stat

aniked [119]

Answer:

The pressure at this depth is $1.235\cdot P_{atm}$ .

Explanation:

According to the statement, the uncompressed fluid stands at atmospheric pressure. By Boyle's Law we have the following expression:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$V_{1}, V_{2}$ - Initial and final volume.

$P_{1}, P_{2}$ - Initial and final pressure.

If we know that $V_{2} = 0.81\cdot V_{1}$ , then the pressure ratio is:

$\frac{P_{2}}{P_{1}} = 1.235$

If $P_{1} = P_{atm}$ , then the final pressure of the gas is:

$P_{2} = 1.235\cdot P_{atm}$

The pressure at this depth is $1.235\cdot P_{atm}$ .

6 0

3 years ago

I need help answering this question.

Sloan [31]

The work done is the loss of kinetic energy.

Loss of kinetic energy = m*(v1^2 - v2^2)/2 = 10 kg * [ (99m/s)^2 - (1m/s)^2]/2 = 49,000 J

6 0

3 years ago

What happens when a negatively charged object A is brought near a neutral object B?

Eva8 [605]

<span>Object B stays neutral but becomes polarized.

Hope this helps.

~Jurgen</span>

5 0

3 years ago

Read 2 more answers