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faust18 [17]
3 years ago
9

Stars within the halo move in random orbits that extend above and below the disk of the galaxy, while stars in the disk move in

orderly, elliptical orbits within the galactic plane. as seen from the outside of the milky way, which of the images below correctly illustrates the orbital motions and distribution of stars within the halo?
Physics
1 answer:
Blizzard [7]3 years ago
4 0
<span>Answer: Spherical Distribution Feedback: Correct The stars in the halo component have highly-inclined random orbits that orbit the center of our Galaxy. The stars within the halo would therefore make up a spherical distribution of stars surrounding the center of the Galaxy. In comparison, the disk stars move in elliptical orbits, which are nearly circular and are confined to the disk of the Galaxy. Disk stars therefore have very small inclinations and do not move above or below the plane of the Galactic disk.</span>
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A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s,
Art [367]

Answer:

\dfrac{d\theta}{dt} =-0.233\ rad/s

Explanation:

given,

length of ladder = 10 ft

let x be the distance of the bottom and y be the distance of the top of ladder.

x² + y² = 100

differentiating with respect to time we get

2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

16\times 1.4 + 12\dfrac{dy}{dt} = 0

\dfrac{dy}{dt} = -\dfrac{5.6}{3}

let the angle between the ladders be θ

tan\theta = \dfrac{y}{x}

y = xtan θ

\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}

-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}

\dfrac{25}{2} \dfrac{d\theta}{dt} =\dfrac{-17.5}{6}

\dfrac{d\theta}{dt} =-0.233\ rad/s

6 0
3 years ago
What does displacement describe?
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3 years ago
If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the mini
kondaur [170]
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3 0
3 years ago
1. You wish to heat 20 kg of water from 40°C to 80°C. How many kcal of heat are necessary to do this? To how many kJ does this c
adelina 88 [10]

Answer:

<h2>3,343.68kJ </h2>

Explanation:

Heat energy used up can be calculated using the formula:

H = mcΔt

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c = specific heat capacity of water = 4179.6J/kg°C

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H= 20* 4179.6 * 40

H = 3,343,680Joules

H = 3,343.68kJ

8 0
3 years ago
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