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3 years ago

Which of the following is NOT an example of general characteristic of expository writing? Select all that apply.. .

Physics

2 answers:

igor_vitrenko [27]3 years ago

7 0

Answer:

1) Thoroughness

2) Persuasiveness

Explanation:

Thoroughness: A thorough writing is one that goes deep into details and reveals much about a topic, this is not an example of expository writing because expository writing only reveals brief information about a topic.

Persuasiveness: A persuasive writing is one with a target to sell an idea or convince someone to buy into the content of a topic, this is not a form of expository writing because expository writing only introduces a topic and does not in anyway persuade a listener to take action.

MA_775_DIABLO [31]3 years ago

5 0

The correct answer should be poetic description

Poetic descriptions are found in other types of essays while expository essays are used to convey information.

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Answer:

12s

Explanation:

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3 years ago

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3 years ago

A rock is thrown downward at a speed of 8.5 m/s and hits the ground 2.25 seconds later. From what height was the rock thrown? Wh

vekshin1

Answer:

The velocity and height is 31 m/s and 44.4 m respectively .

Explanation:

Given :

Initial speed of rock , u = 8.5 m/s .

Time taken to reach ground , t = 2.25 s .

Also , acceleration due to gravity is , $g=10\ m/s^2$ .

Now , to find the velocity when it hit ground .

Applying equation of motion :

$v=u+at\\v=8.5+10\times 2.25\\v=31\ m/s$

Also , height from which rock is thrown is given by :

$h=ut+\dfrac{at^2}{2}\\\\h=8.5\times 2.25+\dfrac{10\times 2.25^2}{2}\\\\h=44.4\ m$

Hence , this is the required solution .

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3 years ago

two astronauts are taking a spacewalk outside the International Space Station the first astronaut has a mass of 64 kg the second

Fittoniya [83]

Answer:

Approximately $0.88\; {\rm m \cdot s^{-1}}$ to the right (assuming that both astronauts were originally stationary.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Since momentum of this system (of the astronauts) conserved:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\end{aligned}$ .

Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be $0$ since the velocity of both astronauts was $0\!$ .

Therefore:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

The final momentum of the first astronaut ( $m = 64\; {\rm kg}$ , $v = 0.8\; {\rm m\cdot s^{-1}}$ to the left) would be $p_{1} = m\, v = 64\; {\rm kg} \times 0.8\; {\rm m\cdot s^{-1}} = 51.2\; {\rm kg \cdot m \cdot s^{-1}}$ to the left.

Let $p_{2}$ denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be $(p_{1} + p_{2})$ .

$\begin{aligned} & p_{1} + p_{2} \\ &= (\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

Hence, $p_{2} = (-p_{1})$ . In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude ( $51.2\; {\rm kg \cdot m \cdot s^{-1}}$ ) but opposite in direction (to the right versus to the left.)

Rearrange the equation $p = m\, v$ to obtain an expression for velocity in terms of momentum and mass: $v = (p / m)$ .

$\begin{aligned}v &= \frac{p}{m} \\ &= \frac{51.2\; {\rm kg \cdot m \cdot s^{-1}}}{64\; {\rm kg}} && \genfrac{}{}{0}{}{(\text{to the right})}{} \\ &\approx 0.88\; {\rm m\cdot s^{-1}} && (\text{to the right})\end{aligned}$ .

Hence, the velocity of the astronaut in question ( $m = 58.2\; {\rm kg}$ ) would be $0.88\; {\rm m \cdot s^{-1}}$ to the right.

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2 years ago