The Universe became dominated by matter instead of radiation at a redshift of around 3300.
To determine at what redshift the Universe became dominated by matter, we need to find the redshift at which the energy density of matter becomes equal to the energy density of radiation.
Let's start with the energy density of radiation, which can be calculated using the Stefan-Boltzmann law:
$[tex]u_{rad} = \frac{4\sigma}{c}T^4$[/tex]
where $\sigma$ is the Stefan-Boltzmann constant, $c$ is the speed of light, and $T$ is the temperature of the radiation. Since we are assuming that the cosmic microwave radiation is a black-body radiation, we can use the temperature of 2.73 K, which corresponds to the peak of the radiation curve:
[tex]$u_{rad} = \frac{4\sigma}{c}(2.73K)^4 \approx 0.261 \text{ eV/cm}^3$[/tex]
Next, let's calculate the energy density of matter. We know that the number density of baryons is [tex]$n_b \approx \frac{1}{10^9}n_{\gamma}$, where $n_{\gamma}$[/tex] is the number density of photons. Since we are assuming that the photon-to-baryon ratio is constant, we can write:
[tex]$\frac{\rho_b}{\rho_{\gamma}} = \frac{m_b n_b}{\frac{4}{3}\sigma T^4} = \frac{3m_b}{4\sigma T^3 n_{\gamma}} \approx \frac{3m_b}{4\sigma T^3}\frac{1}{n_{\gamma}}$[/tex]
where $m_b$ is the mass of a baryon. Substituting the values, we get:
[tex]$\frac{\rho_b}{\rho_{\gamma}} \approx 4.15 \times 10^{-10}$[/tex]
Since the total energy density of the Universe is given by:
[tex]$\rho_{tot} = \rho_b + \rho_{\gamma}$[/tex]
we can write:
[tex]$\frac{\rho_b}{\rho_{tot}} = \frac{\rho_b}{\rho_b + \rho_{\gamma}} \approx \frac{\rho_b}{\rho_{\gamma}} = 4.15 \times 10^{-10}$[/tex]
At the redshift $z$, the energy density of radiation will be diluted by a factor of $[tex](1+z)^4[/tex]$, while the energy density of matter will be diluted by a factor of $[tex](1+z)^3[/tex]$. Thus, at some redshift $z$, we will have:
$ [tex]\frac{\rho_b}{\rho_{tot}} = \frac{\rho_b}{\rho_b + \rho_{\gamma}} = \frac{1}{1+z}\frac{3m_b}{4\sigma T^3 n_{\gamma}}[/tex] $
Setting this equal to the value we calculated above, we can solve for $z$:
$ [tex]\frac{1}{1+z}\frac{3m_b}{4\sigma T^3 n_{\gamma}} \approx 4.15 \times 10^{-10}[/tex] $
$ [tex]1+z \approx \frac{3m_b}{4\sigma T^3 n_{\gamma}}\frac{1}{4.15 \times 10^{-10}}[/tex] $
$ [tex]z \approx 3300[/tex] $
Therefore, the Universe became dominated by matter instead of radiation at a redshift of around 3300.
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can you please tell me where does 1-14 i really need help thanks :) god bless you all
The above has to do with the study of the earth's lithospheric plates. See the attached image and the explanation below.
What are the processes of the movement of lithospheric plates?The movement of lithospheric plates is a geological process that occurs due to the motion of hot, molten material in the Earth's mantle. The lithosphere, which is the rigid outer layer of the Earth's surface, is divided into several large plates that move relative to each other.
These movements are caused by the convection of material in the mantle and the forces that arise at the boundaries between the plates.
There are three main types of plate boundaries: divergent, convergent, and transform. Divergent boundaries occur where plates move apart from each other, creating new oceanic crust. Convergent boundaries arise where plates collide, leading to subduction, volcanic activity, and the formation of mountains. Transform boundaries occur where plates slide past each other.
The movement of lithospheric plates gives rise to various geological phenomena, such as earthquakes, volcanic activity, and the formation of mountain ranges and ocean basins.
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A light ray passing through air strikes the surface of a glass block (n=1.5) and makes 30° angle of incidence. How many degrees will the light ray deviate from its original path after refraction?
The light ray will deviate from its original path with 19.5° after refraction.
How do we calculate?Applying Snell's law to calculate the angle of refraction:
n1 sin θ1 = n2 sin θ2
where n1 and θ1 = the refractive index and the angle of incidence in the first medium (air),
n2 and θ2 = the refractive index and the angle of refraction in the second medium (glass).
In this example,
n1 = 1.00 (refractive index of air), θ1 = 30°, and
n2 = 1.5 (refractive index of glass).
We then calculate for θ2:
n1 sin θ1 = n2 sin θ2
1.00 * sin 30° = 1.5 * sin θ2
0.5 = 1.5 * sin θ2
sin θ2 = 0.5 / 1.5 = 1/3
θ2 = sin^-1(1/3)
θ2 = 19.5°
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30 POINTS!!!! NO CHATGPT OR ANY BOTS_
As you sit in a fishing boat, you notice that 12 waves pass the boat every 45 s
. If the distance from one crest to the next is 9.0 m
, what is the speed of these waves?
Express your answer to two significant figures and include the appropriate units.
The speed of the waves can be expressed to two significant figures as 0.2 m/s. The unit for this expression is meters per second (m/s).
What is wave crest?A wave crest is the highest point of a wave. It is the top of the wave, where the wave is moving most up and away from the equilibrium position. It is the point of highest amplitude (height) of the wave and is followed by a wave trough, which is the lowest point of the wave.
The speed of the waves can be calculated using the formula speed = distance over time.
We know the distance between wave crests is 9.0 m and the time it takes for 12 waves to pass the boat is 45 s. Therefore, the speed of the waves can be calculated as:
Speed = 9.0 m / 45 s
Speed = 0.2 m/s
The speed of the waves can be expressed to two significant figures as 0.2 m/s. The unit for this expression is meters per second (m/s).
This calculation shows that the speed of the waves passing the boat is 0.2 m/s. This speed can be further broken down into how many meters the waves travel in one second if necessary.
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5. Two equal charges are situated in a vacuum 10.0cm apart, if they repel each other with a force of 0.5N, calculate the value of the charge on each. [4π)¹ = 9.0 x 10⁹ I
The value of the charge on each particle is [tex]1.05 x 10^-8 C[/tex].
What is Coulomb's law?Coulomb's law is a fundamental principle of electrostatics that describes the interaction between electric charges. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. We can use Coulomb's law to solve this problem. Mathematically,
[tex]F = k(q1q2)/r^2[/tex]
where F is the force of attraction or repulsion between the two charged particles,[tex]q1[/tex] and [tex]q2[/tex] are the magnitudes of the charges on the two particles, r is the distance between them, and k is Coulomb's constant, which has a value of [tex]9.0 x 10^9 Nm^2/C^2.[/tex]
In this problem, we know that the charges are equal and the distance between them is 10.0 cm. We also know that the force between them is 0.5 N. Therefore,
[tex]0.5 N = k(q^2)/(0.1 m)^2[/tex]
Solving for q, we get:
[tex]q = \sqrt{[(0.5 N)(0.1 m)^2/k]}[/tex]
[tex]q = \sqrt{(0.5 N)(0.01 m)/(9.0 x 10^9 Nm^2/C^2)}[/tex]
[tex]q = 1.05 x 10^-8 C[/tex]
Therefore, the value of the charge on each particle is [tex]1.05 x 10^-8 C.[/tex]
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How long does it take for radiation from a cesuim-133 atom to complete 1.5 million cycles
A cesium-133 atom's radiation goes through 1.5 million cycles in around 0.1633 microseconds (or 163.3 nanoseconds).
What frequency does one kind of radiation that cesium-133 emits have?9,192,631,770 hertz (cycles per second) is the frequency of the microwave spectral line that the isotope cesium-133 emits. The basic unit of time is provided by this. Cesium clocks have an accuracy and stability of 1 second in 1.4 million years.
The radiation emitted by cesium-133 has a frequency of 9,192,631,770 cycles per second, or 9.192631770 109 Hz.
The following formula may be used to determine how long 1.5 million radiation cycles take to complete:
Time is equal to the frequency of cycles.
Plugging in the numbers, we get:
time = 1.5 million / 9.192631770 × 10^9 Hz
time = 1.632995101 × 10^-7 seconds
So it takes approximately 0.1633 microseconds (or 163.3 nanoseconds) for radiation from a cesium-133 atom to complete 1.5 million cycles.
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A model rocket blast off and moves upward with an acceleration of 12m/s2 until it reaches a height of 26m, at which point its engine shuts off and it continues its flight in free fall.
a) What is the maximum height attained by the rocket?
b) What is the speed of the rocket just before it hits the ground?
c) What is the total duration of the rocket's flight?
The attractive electric force between the point charges q and −2q has a magnitude of 2.2 N when the separation between the charges is 1.4 m . k=8.99×109N⋅m2/C2
What is the magnitude of charge q?
The electric force between two point charges is given by the equation
[tex]F=k*q_1*q_2/r^2[/tex]
What is force?The interaction between two things is measured by the physical quantity known as force. It is a vector quantity, and the sign F is frequently used to denote it. When an object interacts with another object, it feels a push or a pull.
where r is the distance between the charges, q1 and q2 are their magnitudes, and k is the Coulomb constant.
When we enter the problem's specified values, we obtain
[tex]2.2N=8.99*10^9\ N*m^2/C^2*q*-2q/(1.4 m)^2[/tex]
which simplifies to
q = -0.500 N/C.
Thus, the magnitude of charge q is 0.500 N/C.
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How loud in Decibels would a sound be with an intensity of 7.8x10^-4 W/m2? (write your answer to one decimal space)
A sound that is 7.8x10-4 W/m2 in intensity is equal to (10 dB)log3.2106 W/m21012 W/m2=185 dB.
How can you determine the relative volume of a sound?The decibel, often known as the db or 0.1 bel, is the standard measurement unit. Hence, b = 10 log10 (I/I0) can be used to express the relationship between relative intensities, or b, in decibels. This equation can be used to determine that one decibel equals a 26 percent intensity variations.
What does physics mean by relative intensity?The "decibel level" of a sound is a less formal term for relative intensity level. It is not the same as energy; relative intensity level reflects loudness more faithfully by using a logarithmic scale.
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How can you determine the number of neutrons in an atom?
A. Mass number plus number of electrons
B. Atomic number minus mass number
C. Mass number minus atomic number
D. Atomic number plus mass number
Answer:
B. Atomic number minus mass number
Explanation:
If the speed of a wave is 400 cm/s with a frequency of 80 Hz, what is the wavelength for this wave?
32,000 cm
32,000 m
5 cm
5m
an election of mass 9.1 × 10^31kg moves with a velocity of 4.2 × 10^7mJs between the cathode and anode of an X-ray tube. Calculate the wavelength.( take Planck's constant, h= 6.6 × 10^ 34 J's)
The wavelength of the electron is 1.724 × 10^-12 m.
How do we calculate?The wavelength of the electron is found using the de Broglie wavelength formula:
λ = h / p
where λ = wavelength,
h= Planck's constant, a
p = momentum of the electron.
we find the momentum of the electron,
p = m * v
p = (9.1 × 10^-31 kg) * (4.2 × 10^7 m/s)
p = 3.822 × 10^-22 kg m/s
Therefore, wavelength ;
λ = h / p
λ = (6.6 × 10^-34 J s) / (3.822 × 10^-22 kg m/s)
λ = 1.724 × 10^-12 m
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Which of the following best describes the relationship between the two variables?
Answer:
correlation is defined as the statistical association between two variables
Explanation:
a correction exits between two variables when one of them is related to the other in some way
As a 5.00-kg sample of liquid mercury is cooled into a solid, it liberates 157 kJ of energy. What is the original temperature of the mercury? For mercury, the melting point is 234 K, the heat of fusion is 11.3 kJ/kg,
and the specific heat is 140 J/kg . K.
378 K
690 K
157 K
410 K
The original temperature of the mercury is 260.6K
Here is how to arrive at temperature of the mercuryTo solve this problem, we can use the formula for the heat released during the solidification of a substance:
Q = m * Lf
where Q is the heat released, m is the mass of the substance, and Lf is the heat of fusion of the substance.
In this case, Q = 157 kJ, m = 5.00 kg, and Lf = 11.3 kJ/kg.
We also need to use the formula for the heat absorbed or released during a temperature change:
Q = m * c * ΔT
where Q is the heat absorbed or released, m is the mass of the substance, c is the specific heat of the substance, and ΔT is the change in temperature.
We can use this formula to calculate the heat released as the mercury cools from its original temperature to its melting point, and then use the formula for solidification to calculate the heat released as the mercury solidifies.
Let T be the original temperature of the mercury.
The heat released as the mercury cools from its original temperature to its melting point is:
Q1 = m * c * (T - 234)
The heat released as the mercury solidifies is:
Q2 = m * Lf
The total heat released is:
Q = Q1 + Q2 = m * c * (T - 234) + m * Lf
Substituting the values given in the problem, we get:
157 kJ = 5.00 kg * 140 J/kg . K * (T - 234) + 5.00 kg * 11.3 kJ/kg
Simplifying and solving for T, we get:
T = 260.6 K
Therefore, the original temperature of the mercury was 260.6 K.
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If the sun were more massive, what would happen to Earth’s gravity with the sun?
A. decrease
B. would be infinite
C. would be 0
D. increase
Answer: d. increase
Explanation:
If the sun were more massive, the gravitational force between the sun and Earth would increase. This means that Earth's gravity with the sun would also increase. Therefore, the correct answer is (D) increase.
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. So, if the mass of one of the objects increases, the gravitational force between them will also increase. In this case, if the mass of the sun were to increase, the gravitational force between the sun and Earth would become stronger, and hence, Earth's gravity with the sun would also increase.
A rock climber stands on top of a 59 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The initial speed of the first stone was 1.7 m/s . Include value and units.
a) How long after the release of the first stone does the second stone hit the water?
b) What was the initial speed of the second stone?
c) What is the speed of the first stone as it hits the water?
d) What is the speed of the second stone as it hits the water?
a) The time after the release of the first stone that the second stone hits the water is 2.0 s.
b) 15.7 m/s is the initial speed of the second stone.
c) The speed of the first stone as it hits the water is 15.7 m/s.
d) The speed of the second stone as it hits the water is 28.2 m/s.
What is velocity?Velocity is a vector quantity that measures both the speed and direction of an object's motion. It is equal to the rate of change of an object's position with respect to time. Velocity is usually represented by the symbol v and is measured in meters per second (m/s).
a) The time between first and second stone's release is 1.0 s. Since the time of release of first stone and the time of splash of both stones are same, the time between the release of second stone and the splash of both stones is 1.0 s.
Thus, the time after the release of the first stone that the second stone hits the water is 2.0 s.
b) The initial speed of the second stone can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (1.7)² + 2(9.8) * 59
v = 15.7 m/s
c) The speed of the first stone as it hits the water can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (1.7)² + 2(9.8) * 59
v = 15.7 m/s
d) The speed of the second stone as it hits the water can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (15.7)² + 2(9.8) * 59
v = 28.2 m/s
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According to this graph, the acceleration
is approximately:
A. 12 m/s²
C. 4 m/s²
Velocity (m/s)
14
12
10
12 2 3 4
Time t (s)
B. 1.5 m/s2
D. 3 m/s2
Help please
Answer:
Explanation:
Because you have velocity along the y axis and time along the x axis, this is a velocity v time graph which is an acceleration graph. The slope of the line in this graph IS the acceleration. We can use 2 points and the slope formula to solve for the acceleration:
(0, 0) and (1, 3):
[tex]m=\frac{3-0}{1-0}=3[/tex] m/s squared, choice D.
A 0.80kg block of carbon (solid) is dropped into 1.4kg of water. If the carbon starts at -20C, the water starts at 92C, and they have equal final temperatures, what is the final temperature of the system?
The system's final temperature is roughly 16.7°C.
What is a system's final temperature?You may determine your substance's final heat by multiplying the temperature change by the initial temperature. Your water's final temperature would be 24 + 6, or 30 degrees Celsius, for instance, if it started off at 24 degrees Celsius.
The following is the formula for energy conservation:
Q1 + Q2 = 0
Q = mcΔT
Q1 + Q2 = 0
568.8
Simplifying and solving for
6394.4 - 106768 = 0
= 16.7°C
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