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Eduardwww [97]
3 years ago
10

When broadcasting live on social, keep in mind that the best broadcasts are ones that feel like a conversation between brand and

viewer. Unlike other forms of social video, you’ll get more views and engagement if your video ________.a. is longer and repeats key points.b. is full of interesting transitions and editing techniques.c. uses unusual lighting and sound effects.d. includes a pdf attachment with details from the broadcast.e. is short and action packed.
Physics
1 answer:
Andru [333]3 years ago
6 0

Answer:

When broadcasting live on social, keep in mind that the best broadcasts are ones that feel like a conversation between brand and viewer. Unlike other forms of social video, you’ll get more views and engagement if your video

is longer and repeats key points.

Explanation:

When broadcasting live on social media, one should be live for long because in this way one can get more views as audience will increase with time. There should be an interaction with the audience like answering their questions which they write in the comments section. These comments and views will make this video to the top of news feed. Secondly the most important thing is the content of the video. One must focus on the information or knowledge he/she wants to convey and must repeat the key points again and again so that one who has missed the important points will be able catch them later.

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When the atom's electrons step down to lower energy levels in a thin cloud of hot gas, what is produced?
Reika [66]

Answer: The correct answer is an emission line spectrum.

Explanation:

When the electrons are excited to the higher energy level, the energy is absorbed in this case.

When the electrons in the atom step down to lower energy levels in a thin cloud of hot gas then the radiation will emit.

The electron will lose energy in this case in the form of radiation. There will be an emission line spectrum.

8 0
3 years ago
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What is the gravitational force of attraction between a planet and a 17-kilogram mass that is falling freely toward the surface
PolarNik [594]

Answer:

a. 150 N

Explanation:

Gravitational Force: This is the force that act on a body under gravity.

The gravitational force always attract every object on or near the earth's surface. The earth therefore, exerts an attractive force on every object on or near it.

The S.I unit of gravitational force is Newton(N).

Mathematically, gravitational force of attraction is expressed as

(i) F = GmM/r² ........................ Equation 1 ( when it involves two object of different masses on the earth)

(ii) F = mg ............................... Equation 2 ( when it involves one mass and the gravitational field).

Given: m = 17 kg, g = 8.8 m/s²

Substituting into equation 2,

F = 17(8.8)

F = 149.6 N

F ≈ 150 N.

Thus the gravitational force = 150 N

The correct option is a. 150 N

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3 years ago
What is NOT one of the three primary resources that families have to reach financial goals?
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What is NOT one of the three primary resources that families have to reach financial goals? It is c) education
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A 905 kg test car travels around a 3.25 km circular track. if the magnitude of the force that maintains the car's circular motio
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Suppose that a spiral galaxy is located at the center of a spherically symmetric dark matter halo
Novosadov [1.4K]

To solve the problem it is necessary to use the concepts of Orbital Speed considering its density, and its angular displacement.

In general terms the Orbital speed is described as,

V_{orbit} = \sqrt{\frac{G\rho 4\pi r^3}{3}}

PART A) If the orbital speed of a star in this galaxy is constant at any radius, then,

\frac{4\pi G\rho r}{3} = \frac{v^2}{r}

\frac{4\pi G\rho r}{1} = \frac{3v^2}{r}

\frac{\rho}{1} = \frac{3v^2}{r^2 4\pi G}

\rho = \frac{1}{r^2}

PART B) This time we havev=\omega t, where \omega is the angular velocity (constant at this case)

\frac{4\pi G\rho r}{3} = \frac{v^2}{r}

\frac{4\pi G\rho r}{3} = \frac{(\omega r)^2}{r}

\rho = \frac{3\omega r}{4\pi Gr}

\rho = \frac{3\omega^2}{4\pi G} \propto constant

PART C) If the total mass interior to any radius r is a constant,

\frac{4\pi G\rho r}{3} = \frac{v^2}{r}

\frac{GM}{r^2}=\frac{v^2}{r}

v = \sqrt{\frac{GM}{r}}

v= \sqrt{\frac{1}{r}}

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4 years ago
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