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steposvetlana [31]
1 year ago
5

the water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of

the hose is pointed straight up. the tank cover is airtight, and the air pressure above the water surface is 2 atm gage. the system is at sea level. determine the maximum height to which the water stream could rise
Engineering
1 answer:
Brilliant_brown [7]1 year ago
4 0

The maximum height of the water stream that could rise is 40.65 m. It can be calculated using Principle of Bernoulli.

Principle of Bernoulli can be described an increase in the speed of a fluid happen simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. Equation of Bernoulli is shown as a conservation of energy law for a flowing fluid. Bernoulli equation is shown below:

P_1/ρg + v_1²/2g + h_1  = P_2/ρg +v_2²/2g + h_2

Where:

ρ = fluid density

g = acceleration due to gravity

P_1 = pressure at elevation 1

v_1 = velocity at elevation 1

h_1 = height of elevation 1

P_2 = pressure at elevation 2

v_2 = velocity at elevation 2

h_2 = height at elevation 2

Assumed that the water at free surface, so v_1 = v_2 = 0

So, the formula will be

P_1/ρg + h_1  = Patm/ρg + h_2

Based on the scenario, we know that:

h_1 = 20 m

P_1gage = 2 atm ≈ 20265 N/m

ρ = 1000 kg/m²

From the scenario, we will determine the maximum height by using the equation of bernoulli and it will be

h_2 = (P_1 - Patm)/ρg + Z_1

h_2 = \frac{2 atm}{1000(9.81)} x (\frac{101325N/m^{2} }{1 atm} )(\frac{1 kg.m/s^{2} }{1 N} ) + 20 = 40.65 m

So, from that, we can conclude the maximum height of the water stream that could rise is 40.65 m.

Learn more about bernoulli principle at brainly.com/question/15415820

#SPJ4

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Consider three charged sheets, A, B, and C. The sheets are parallel with A above B above C. On each sheet there is surface charg
babunello [35]

Answer:

0.

Explanation:

To find the electrical force per unit area on each sheet we start defining our variables,

\sigma_A = -4.10^{-5}C/m^2

\sigma_B= -7.10^{-5}C/m^2

\sigma_C = -3.1^{-5}C/m^2

We find the electric field for each one, this formula is given by,

E= \frac{\sigma_i}{2\epsilon_0}

Substituting each value from the three charged sheets, we have

E_A= \frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})

E_B= \frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})

E_C= \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})

The electric field is

E_{NET}= E_A+E_B+E_C

E_{NET}=\frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})+\frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})+ \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})

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5 0
3 years ago
The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
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Answer:

a) 254.6 GPa

b) 140.86 GPa

Explanation:

a) Considering the expression of rule of mixtures for upper-bound and calculating the modulus of elasticity for upper bound;

Ec(u) = EmVm + EpVp

To calculate the volume fraction of matrix, 0.63 is substituted for Vp in the equation below,

Vm + Vp = 1

Vm = 1 - 0.63

Vm = 0.37

In the first equation,

Where

Em = 68 GPa, Ep = 380 GPa, Vm = 0.37 and Vp = 0.63,

The modulus of elasticity upper-bound is,

Ec(u) = EmVm + EpVp

Ec(u) = (68 x 0.37) + (380 x 0.63)

Ec(u) = 254.6 GPa.

b) Considering the express of rule of mixtures for lower bound;

Ec(l) = (EmEp)/(VmEp + VpEm)

Substituting values into the equation,

Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)

Ec(l) = 25840/183.44

Ec(l) = 140.86 GPa

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Answer:

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its B

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