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svlad2 [7]
3 years ago
5

Water is pumped from a reservoir to a tank suspended 12 m above the reservoir. The water flows through the same size pipe throug

hout the process. Both the reservoir and the tank are open to the atmosphere. If friction in the pipes consumes 20% of the pump energy, determine the mass flow of water if a 2 hp pump is used. Assume the liquid water has a density of 1 g/cm3

Physics
1 answer:
Nady [450]3 years ago
8 0

Answer:

Explanation:

the solution is given in the pictures attached below

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Two people walking on a sidewalk have the following
vova2212 [387]

Answer:

X2 is fasteer

x=0 will go to Xi

Explanation:

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Which of the following is an ionic compound ? A. Gold
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Among the given choices, the ionic compound is D. Magnesium Chloride (<span>MgCl2</span>) where magnesium has a +2 charge while chloride has a -1 charge. Ionic compounds are chemical compounds comprising of ions held together by electrostatic forces named as ionic bonding.
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How can you use tweezers to separate a mixture
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Hi, this sounds like a chemistry question:
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3 years ago
Read 2 more answers
A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a d
Snowcat [4.5K]

Answer:

The energy stored is 1.4 x 10^-9 J.

Explanation:

Side of square, L = 10 cm = 0.1 m

Distance, d = 2 mm = 0.002 m

Electric field, E = 4000 V/m

The energy stored in the capacitor is

U = 0.5 C V^2

The capacitance is given by

C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J

7 0
3 years ago
A total charge of Q coulomb is uniformly distributed along a rod 40cm in length.find find the electric field intensity 20cm away
Tatiana [17]

For a total charge of Q coulomb is uniformly distributed along a rod 40cm in length,  the electric field intensity 20cm away from the rod is mathematically given as

E1=1.598*10^11v/m

<h3>What is the electric field intensity 20cm away from the rod along its perpendicular bisector?</h3>

Generally, the equation for the  initial electric field intensity   is mathematically given as

dEp=\int{kd/r}cosd\theta

Therefore

e1=kd/r{sin\theta2+sinR1}

Hence

E1=(9*10^9/20*10^{-2})({sin45+sin45})*B/40*10^-2

E1=B*9*10^{13})/(10*110)*\sqrt{2}

E1=1.598*10^11v/m

In conclusion, the electric field intensity

E1=1.598*10^11v/m

Read more about Electric field

brainly.com/question/9383604

6 0
2 years ago
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