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Likurg_2 [28]
3 years ago
8

The author s feeling about a subject or topic, which is evidenced in word choice, is called __________. conflict resolution tone

tension
Physics
2 answers:
Stels [109]3 years ago
6 0
The author s feeling about a subject or topic, which is evidenced in word choice, is called tone.
xeze [42]3 years ago
3 0
<h3><u>Answer;</u></h3>

Tone

The author s feeling about a subject or topic, which is evidenced in word choice, is called <u>tone</u>.

<h3><u>Explanation</u>;</h3>
  • <em><u>Tone indicates or describes the attitude of the author towards a given subject or topic. </u></em>The author's attitude is expressed through the choice of words he or she uses.
  • <u><em>The tone is described by adjectives such as ironic, cynical, intense, sarcastic etc. It may be expressed by use of syntax, point of view, diction, etc. </em></u>
  • <u><em>The tone may change very quickly or may remain the same throughout a topic or a story.</em></u>
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how far can a mother push a 20.0kg baby carriage, using a force of 62.0N at angle of 30.0°to the horizontal, if she can do 2920
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Here is my solution using my app.

4 0
3 years ago
A push of magnitude P acts on a box of weight W as shown in the figure. The push is directed at an angle θ below the horizontal,
Kipish [7]

Answer:

its on wheels and they are supposed to make it eas

Explanation:

3 0
2 years ago
A ball is thrown at 60 degrees and lands in 18.5 seconds what is the velocity of the ball at the start
n200080 [17]

Answer:

the initial velocity of the ball is 104.67 m/s.

Explanation:

Given;

angle of projection, θ = 60⁰

time of flight, T = 18.5 s

let the initial velocity of the ball, = u

The time of flight is given as;

T = \frac{2u\times sin(\theta)}{g} \\\\2u\times sin(\theta) = Tg\\\\u = \frac{Tg}{2\times sin(\theta)} \\\\u = \frac{18.5 \times 9.8}{2\times sin(60^0)} \\\\u = 104.67 \ m/s

Therefore, the initial velocity of the ball is 104.67 m/s.

3 0
3 years ago
During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying
Jlenok [28]

Well, I guess you can come close, but you can't tell exactly.

It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.

That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.

  98.5 m/s  =  √ [ (horizontal component)² + (vertical component)² ].

The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:

             Height = (1/2) (acceleration) (time²) .

If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.

                  Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
                 Height = (1/2) (9.81) (10.04)²

                            =   (4.905 m/s²) x (100.8 sec²)  =  494.43 meters.

As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.

If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component.  That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.   

8 0
3 years ago
Plz answer this asap no explanation needed answer both questions first answer gets brainliest
Soloha48 [4]

Answer:

8) d

9) c

<em>Hope this helps! :)</em>


5 0
3 years ago
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