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3 years ago

12

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

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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2 years ago

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jonny [76]

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3 years ago

Can someone answer this question?<br><br><br><br><br><br><br> {For me: question 48}

qaws [65]

0.000000218 = 2 x 10^ -8

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3 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him

juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal direction

$m_{b}v_{b}\cos\theta= m_{c}v_{c}$

$v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}$

Put the value into the formula

$v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}$

$v_{c}=0.0345\ m/s$

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0

3 years ago

A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across

NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:

$A=\frac{\pi }{4}d^{2} \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\$

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>

<em> wire: </em>

L= 2.00 m

From Table Copper Resistivity $p$ = 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

$R=\frac{pL}{A}$

=0.0165Ω

The Potential difference across the copper wire is:

V=IR

=2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:

$R=\frac{pL}{A}$

=0.014Ω

The Potential difference across the Silver wire is:

V=IR

=1.75*10^-4

4 0

3 years ago