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3 years ago

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Detailed measurements of the disk and central bulge region of our Galaxy suggest our Milky Way is a:Select one:A. quasar. B. bar

red spiral galaxy. C. very flat elliptical galaxy. D. very dusty irregular galaxy. E. normal spiral galaxy.

1 answer:

chubhunter [2.5K]3 years ago

4 0

I believe the answer is:

~D. Very dusty irregular galaxy.

Hope this helps!!!

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Learning Task 2: Prepare a basin with half-filled water and stone. Drop a stone

ELEN [110]

Answer:

1 . What happens when you drop the stone?

Depending on the weight from which the stone was dropped, the glass might well break

2 depending on the size and weight and shape on the stone the glass might well break

3 depending on the density on the stone the stone might when float on the water

Explanition :

GIVE ME BRAINLESS PLEASE !!

6 0

3 years ago

Guys please helpp!!!!1

Setler79 [48]

Answer:

Position A/Position E

$K = E$ , $U = 0$

Position B/Position D

$K = (1-x)\cdot E$ , $U = x\cdot E$ , for $0 < x < 1$

Position C

$K = 0$ , $U = E$

Explanation:

Let suppose that ball-Earth system represents a conservative system. By Principle of Energy Conservation, total energy ( $E$ ) is the sum of gravitational potential energy ( $U$ ) and translational kinetic energy ( $K$ ), all measured in joules. In addition, gravitational potential energy is directly proportional to height ( $h$ ) and translational kinetic energy is directly proportional to the square of velocity.

Besides, gravitational potential energy is increased at the expense of translational kinetric energy. Then, relative amounts at each position are described below:

Position A/Position E

$K = E$ , $U = 0$

Position B/Position D

$K = (1-x)\cdot E$ , $U = x\cdot E$ , for $0 < x < 1$

Position C

$K = 0$ , $U = E$

3 0

3 years ago

Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th

Akimi4 [234]

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

6 0

3 years ago

A substance that contains only one kind of matter is known as a(n)

Anna35 [415]

An element would be the type of substance that only contains one kind of matter.

3 0

4 years ago

Calculate the magnitude of electric field strength at a point 3cm from an infinite line of charge of linear density 18 μC/cm sit

valina [46]

Answer:

The magnitude of the electric field strength = 7.2 x 10⁸ N/C

Explanation:

The linear density:

$\begin{gathered} \lambda=18\mu C\text{ /cm} \\ \\ \lambda=\frac{18*10^{-6}C}{0.01m} \\ \\ \lambda=0.0018\text{ C/m} \end{gathered}$

Point r = 3 cm = 3/100 m

r = 0.03 m

The electric field strength is calculated below

$\begin{gathered} E=\frac{\lambda}{2\pi\epsilon_o\epsilon_rr} \\ \\ E=\frac{0.0018}{2\times3.14\times8.85\times10^{-12}\times1.5\times0.03} \\ \\ E=719709237.468\text{ N/C} \\ \\ E=7.2\times10^8\text{ N/C} \\ \end{gathered}$

The magnitude of the electric field strength = 7.2 x 10⁸ N/C

8 0

1 year ago