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3 years ago

14

How does the amount of dark matter in clusters of galaxies compare to the amount of visible matter?

1 answer:

faust18 [17]3 years ago

3 0

Thus, dark matter constitutes 85% of total mass, while dark energy plus dark matter constitute 95% of total mass–energy content. Because dark matter has not yet been observed directly, if it exists, it must barely interact with ordinary baryonic matter and radiation, except through gravity.

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Types of energy i really need this answer

anygoal [31]

Answer:

These are the two basic forms of energy. The different types of energy include thermal energy, radiant energy, chemical energy, nuclear energy, electrical energy, motion energy, sound energy, elastic energy and gravitational energy.

Explanation:

8 0

3 years ago

Read 2 more answers

Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .

jek_recluse [69]

Explanation:

Given that,

Force, $F=((-8i)+6j)\ N$

Position of the particle, $r=(3i+4j)\ m$

(a) The toque on a particle about the origin is given by :

$\tau=F\times r$

$\tau=((-8i)+6j) \times (3i+4j)$

Taking the cross product of above two vectors, we get the value of torque as :

$\tau=(0+0-50k)\ N-m$

(b) Let $\theta$ is the angle between r and F. The angle between two vectors is given by :

$cos\theta=\dfrac{r.F}{|r|.|F|}$

$cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }$

$cos\theta=\dfrac{0}{50}$

$\theta=90^{\circ}$

6 0

3 years ago

A certain superconducting magnet in the form of a solenoid of length 0.70 m can generate a magnetic field of 8.0 T in its core w

vampirchik [111]

Answer:

16

Explanation:

8 0

3 years ago

A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms

Aleonysh [2.5K]

Explanation:

It is given that,

Power of EM waves, P = 1800 W

We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.

Intensity,

$I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2$

The formula that is used to find the rms value of the electric field is as follows :

$I=\epsilon_o cE^2_{rms}$

c is speed of light and $\epsilon_o$ is permittivity of free space

So,

$E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m$

Hence, this is the required solution.

4 0

3 years ago

Be sure to show all 4 steps for each of the problems below.

Kisachek [45]

Answer:

1=920

2=2300

Explanation:

from the question,

M¹=23

H¹=4

g=10(constant)

using the formula

P. E,= mgh

P. E= 23×10×4=920

P. E=23×10×10=2300

7 0

4 years ago