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il63 [147K]
3 years ago
14

A volumetric flow rate of 2 m^3/ s of liquid water is to be pumped from sea level to an elevation of 1200 m. Taking the density

of the water is 1000 kg/m^3, calculate the minimum power requirement necessary to accomplish this task.
Physics
1 answer:
exis [7]3 years ago
6 0

Answer:

power requirement is 23.52 × 10^{6}  W

Explanation:

given data

flow rate q = 2 m³/s

elevation h = 1200 m

density of the water ρ = 1000 kg/m³

to find out

power requirement

solution

we will get power by the power equation that is

power = ρ× Q× g× h   ...................1

put here all value we get power

power = ρ× Q× g× h

power = 1000 × 2 × 9.8 × 1200

power = 23.52 × 10^{6}

so power requirement is 23.52 × 10^{6}  W

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UkoKoshka [18]

Data:

u=0 m/s is the initial velocity of the plane

v=62 m/s is the final velocity of the plane (at which the plane takes off)

a=1.7 m/s^2 is the acceleration of the plane

To find the minimum distance S the plane needs to take off, we can use the following equation:

2aS=v^2 -u^2

Re-arranging it and substituting the numbers, we find

S=\frac{v^2-u^2}{2a}=\frac{(62 m/s)^2-0}{2(1.7 m/s^2)}=1131 m

3 0
3 years ago
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enot [183]

Answer:

T=+1.133N

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Tension and weight are forces that have opposite directions

Weight is negative (downward)

W=m*g= 0.11kg*(-9.8m/s^2)

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Ft= T+W

Also the total force is the product of the mass due to acceleration:

Ft=m*a

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Ft=+0.055N (upward)

Tension will be the difference between Ft and W:

T= Ft-W

T=+0.055N-(-1.078N)

T=+1.133N

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steposvetlana [31]
The electric forces on both beads are equal.
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The simplest way to do this is to set up equivalent fractions, like this- 

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Solve for x by using cross multiplication.

40*2.2= 88
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Therefore, the boy weighs 88lbs. 
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Pavel [41]
Hopefully this will help you.

4 0
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