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2 years ago

Why would we expect protogalactic clouds with relatively high density to form an elliptical galaxy rather than a spiral galaxy?

1 answer:

7 0

The higher density allows the protogalactic clouds to cool faster and form an elliptical galaxy rather than a spiral galaxy.

In physical cosmology, a protogalaxy or protogalactic cloud , which could also be called a "primeval galaxy", is a cloud of gas which is forming into a galaxy. It is believed that the rate of star formation during this period of galactic evolution will determine whether a galaxy is a spiral or elliptical galaxy; a slower star formation tends to produce a spiral galaxy. The smaller clumps of gas in a protogalaxy form into stars.

Composition

Since there had been no previous star formation to create other elements, protogalaxies would have been made up almost entirely of hydrogen and helium. The hydrogen would bond to form H2 molecules, with some exceptions. This would change as star formation began and produced more elements through the process of nuclear fusion.

Mechanics

Once a protogalaxy begins to form, all particles bound by its gravity begin to free fall towards it. The time taken for this free-fall to conclude can be approximated using the free-fall equations. Most galaxies have completed this free-fall stage to become stable elliptical or disk galaxies, the disks taking longer to fully form. The formation of galaxy clusters takes much longer and is still in progress now.

This stage is also where galaxies acquire most of their angular momentum. A protogalaxy acquires this due to gravitational influence from neighbouring dense clumps in the early universe, and the further the gas is away from the centre, the more spin it gets.

Learn more about protogalactic clouds here : brainly.com/question/28166070

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The 0.15kg baseball has a speed of v=30 m/s just before it is struck by the bat. It then travels along the trajectory shown befo

notka56 [123]

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

The mass of the baseball, m = 0.15 kg

The speed at which it moves, v = 30 m/s

Time at which the baseball was in contact with the bat, t = 0.75 ms

t = 0.75/1000 s

t = 0.00075 s

The impulsive force is given by the formula:

$F=\frac{mv}{t}$

Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:

$F=\frac{0.15 \times 30}{0.00075} \\\\F=6000N$

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

Learn more here: brainly.com/question/25892144

4 0

2 years ago

Olaf takes a picture of his dog using a digital camera. Which is true regarding the image of the dog detected by sensors in the

trasher [3.6K]

Answer: a. forms where light rays coming from the dog converge.

Explanation:

While photographing through digital camera, the light from the object through the lens converges and falls on image sensor.

The image breaks into many pixels and color and brightness of each pixel is measured and stored in form of number. This forms a digital image.

Thus, the correct option is a because the lens converges the light coming from the object onto the sensor.

6 0

3 years ago

Read 2 more answers

A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh

ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

θ = 273.4 º

b) Kf = 37% of Ko

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved.
Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

$p_{o1x} = 2.00 kg*m/s (1)$

$p_{o1y} = 0 (2)$

We can do the same for the particle moving along the positive y-axis:

$p_{o2x} = 0 (3)$

$p_{o2y} = 4.00 kg*m/s (4)$

Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

$p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)$

$p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)$

Now, the total initial momentum, along these directions, must be equal to the total final momentum.
We can write the equation for the x- axis as follows:

$p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)$

We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

$p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)$

Now, we can repeat exactly the same process for the y- axis, as follows:

$p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)$

We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

$p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)$

Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

$p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)$

We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

$tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)$

The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

$\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)$