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Alexxandr [17]
3 years ago
12

Question 19 of 20

Physics
1 answer:
Alex Ar [27]3 years ago
3 0
They might be an example of Lubricants
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Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record w
Scorpion4ik [409]

Answer:

Explanation:

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.

8 0
3 years ago
The most likely models of the planet Mercury indicate that more than half the planet may be composed of________.
Oxana [17]

Answer:

iron

Explanation:

4 0
3 years ago
Convert 172 to a scientific notation
gtnhenbr [62]

Answer:

1.72×10^2

Explanation:

Is the answer correct ?

6 0
3 years ago
Read 2 more answers
A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How lo
gregori [183]

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

                       H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

 Height at any time 'T' after the drop =

                          (initial height) +

                                              (initial velocity) x (T) +
                                                                 (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

     Upward  = the positive direction

                       Initial height = +150 m
                       Initial velocity = + 3 m/s

     Downward = the negative direction

                     Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

                 H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
0    =  (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

                   -4.9 T²  +  3T  + 150  =  0

Use the quadratic equation:

                         T  =  (-1/9.8) [  -3 plus or minus √(9 + 2940)  ]

                             =  (-1/9.8) [  -3  plus or minus  54.305  ]

                             =  (-1/9.8) [ 51.305  or  -57.305 ]

                          T  =  -5.235 seconds    or    5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

                    H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
H'(t)  =  v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set                                  v₀ + a T  =  0

                                      +3  -  9.8 T  =  0

Add 9.8 to each  side:   3               =  9.8 T

Divide each side by  9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh !  I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
                       (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) =  1.5 m/s .

Distance climbed during that time = (average speed) x (time)

                                                           =  (1.5 m/s) x (0.306 sec)

                                                           =  0.459 meter  (hardly any at all)

     But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at  150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely  5.847 seconds after its release.  

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work.  The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

6 0
4 years ago
A watermelon is dropped from the top of a 80m tall building. We want to find the velocity of the watermelon after it falls for 1
White raven [17]

Answer:

v_f = v_i + at

v_f = 13.23 m/s

Explanation:

Height Of the watermelon when it is dropped is given as

h = 80 m

time of fall under gravity

t = 1.35 s

now if water melon start from rest then we have

v_i = 0

acceleration due to gravity for watermelon

a = 9.81 m/s^2

now we need to find the final speed of watermelon

v_f = v_i + at

so we will have

v_f = 0 + (9.81)(1.35)

v_f = 13.23 m/s

7 0
4 years ago
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