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allsm [11]
3 years ago
7

Poems that focus on one image usually have what purpose? PLEASE HELP MEH!!

Engineering
2 answers:
____ [38]3 years ago
5 0

Answer:

Poems that focus on one image usually have what purpose? PLEASE HELP MEH!!

<em>A. to make readers understand how one event leads to another</em>

B. to make readers look at something in a new and different way

C. to make a point about how two or three things are alike

Explanation:

the one is high lighted

Alinara [238K]3 years ago
5 0

Answer:

A. to make readers understand how one event leads to another

Explanation:

Hope this helped have an amazing day!

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Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 m
lara31 [8.8K]

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

Exit velocity of nozzle = 20m/s

Exit diameter = 40 mm

We know that flow rate Q is given as

Q = A \times V

where A is Area

A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2

Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s

minimum flow rate provided by pump is 0.02513 m^3/s

5 0
3 years ago
Car insurance incentives and discounts are available depending on _____. A. school attendance and driver skill B. vehicle type a
WARRIOR [948]

Answer:D. Location, vehicle type, and driving habits

5 0
3 years ago
Read 2 more answers
A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well
ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity  ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

                                                       = 14.9 * 20.1 = 299.5  m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

  = ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

       = 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = \frac{Q}{4\pi T} * W (u)

 where : Q = 2725.5

               T = 299.5

               W(u)  = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0
3 years ago
Which of the following is NOT an example of a direct cost of workplace injuries?
hammer [34]

Answer:

Lost productivity

4 0
3 years ago
Only answer this if your name is riley
Sati [7]

Answer:

hey im like kinda riley

Explanation:

y u wanna talk to moi

3 0
3 years ago
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