All categories

3 years ago

Unexpected orbital velocities of stars around the centers of galaxies led astronomers to predict the existence of dark _____.

Physics

2 answers:

velikii [3]3 years ago

8 0

Answer:

matter

Explanation:

The stars are orbiting around a matter which is not visible. This matter is called dark matter. They are very massive thus, their force of attraction is very large.

There is lot of dark matter in our universe.

Molodets [167]3 years ago

6 0

<span>Unexpected orbital velocities of stars around the centers of galaxies led astronomers to predict the existence of dark matter. Dark matters are hypothetical substance that are believed to account for around five-sixths of the matter in the universe.</span>

You might be interested in

Two speakers placed 0.94 m apart produce pure tones in sync with each other at a frequency of 1630 Hz. A microphone can be moved

Alina [70]

Answer:

a. approximately $1.1\; \rm m$ (first minimum.)

b. approximately $2.2\; \rm m$ (first maximum.)

c. approximately $3.4\; \rm m$ (second minimum.)

d. approximately $4.7\; \rm m$ (second maximum.)

Explanation:

Let $d$ represent the separation between the two speakers (the two "slits" based on the assumptions.)

Let $\theta$ represent the angle between:

the line joining the microphone and the center of the two speakers, and
the line that goes through the center of the two speakers that is also normal to the line joining the two speakers.

The distance between the microphone and point $P_0$ would thus be $9.4\, \tan(\theta)$ meters.

Based on the assumptions and the equation from Young's double-slit experiment:

$\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}$ .

Hence:

$\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right)$ .

The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let $\lambda$ denote the wavelength of this wave.

$\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}$ .

Calculate the wavelength of this wave based on its frequency and its velocity:

$\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m$ .

Calculate $\theta$ for each of these path differences:

$\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of $\theta$} \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}$ .

In each of these case, the distance between the microphone and $P_0$ would be $9.4\, \tan(\theta)$ . Therefore:

At the first minimum, the distance from $P_0$ is approximately $1.1\; \rm m$ .
At the first maximum, the distance from $P_0$ is approximately $2.2\; \rm m$ .
At the second minimum, the distance from $P_0$ is approximately $3.4\; \rm m$ .
At the second maximum, the distance from $P_0$ is approximately $4.7\;\rm m$ .

6 0

4 years ago

A car accelerates from rest to 37.3 m/s<br> in 7 sec. What is the acceleration?

Sergio039 [100]

Answer:

poopgshshsh

Explanation:

cacawusususu

8 0

3 years ago

What clues should you look for to determine if an authors purpose is to make you think deeply

8090 [49]

If the context is difficult to understand they want you to think hard to get what their trying to say and make you feel a certain way.

7 0

3 years ago

a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t

lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0

3 years ago

The _______of force will determine the _______of acceleration.

loris [4]

ANSWER: the POWER of force will determine the SPEED of acceleration.

EXPLANATION

3 0

4 years ago