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maw [93]
1 year ago
13

According to x-ray observations, the space between galaxies in a galaxy cluster is?

Physics
1 answer:
crimeas [40]1 year ago
6 0

According to x-ray observations, the space between galaxies in a galaxy cluster is very hot. It is because the matter between galaxies (often called the intergalactic medium) is mostly hot, ionized hydrogen with bits of heavier elements such as carbon, oxygen and silicon thrown in.

Massive structures are collapsing than at earlier times. Large collapsing structures lead to higher velocity intergalactic shocks and, as a result, significant intergalactic shock heating, with some gas heated well above the 10^{4} K temperatures.

Heating also occurs as galaxies expel out most of the gas that fell into them. The final product is a warm/hot phase, with temperatures of > 10^{5} K.

Now, Let's know how do you use X-rays to make space observations?

X-radiation is absorbed by the Earth's atmosphere, so instruments to detect X-rays must be taken to high altitude by balloons, sounding rockets, and satellites.

To learn more about Galaxy Cluster, here

brainly.com/question/16557484

#SPJ4

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Which of the following best demonstrates the effect of static friction? A. A person pushing a couch and it slowly sliding across
Allisa [31]
<span> B. A person moving a ball through a stream of water</span>
5 0
2 years ago
What material parameters determine resistivity?
sleet_krkn [62]

Answer:

Resistivity \rho =\frac{RA}{l}

It depends upon cross sectional area and length of material

Explanation:

The resistance of any material is given by R=\frac{\rho l}{A}, here \rho is the resistivity of material , l is length of material and A is cross sectional area

So resistivity \rho =\frac{RA}{l}

So resistuivity of any material depends upon area of cross section and length of material

If cross sectional area will be more then resistivity will be more. And is length of the material will be more then resistivity will be less

3 0
2 years ago
Energy is converted from solar to chemical in Process A and then from one form of chemical to another in Process B. Process A is
Pepsi [2]
I think the answer is photosynthis, when plants turn light into food and energy.
9 0
3 years ago
Read 2 more answers
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
2 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
2 years ago
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