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3 years ago

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Responding to the campaign of 4 classes, 7A, 7B, 7C, 7D contributed the amount of support proportional to the numbers 8,6;7;5 kn

owing the total amount of the two classes 7A, 7B is more than 7D 810 thousand VND sso the money each class has contributed

1 answer:

Zolol [24]3 years ago

8 0

Writing life on my fantasy planet

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What is the difference between the pressure head at the end of a 150m long pipe of diameter 1m coming from the bottom of a reser

uysha [10]

Answer:

$\frac {p_2- p_1}{\rho g} = 31.06 m$

Explanation:

from bernoulli's theorem we have

$\frac{p_1}{\rho g} + \frac{v_1^{2}}{2g} +z_1 = \frac{p_2}{\rho g} + \frac{v_2^{2}}{2g} +z_2 + h_f$

we need to find pressure head difference i.e.

$\frac {p_2- p_1}{\rho g} = (z_1 - z_2) - h_f$

where h_f id head loss

$h_f = \frac{flv^{2}}{D 2g}$

velocity v = $\frac{1}{n} * R^{2/3} S^{2/3}$

$S = \frac{\delta h}{L} = \frac{40}{150} = 0.267$

hydraulic mean radius R = $\frac{A}{P} = \frac{hw}{2h+w}$

$R = \frac{40*1}{2*40+1} = 0.493 m$

so velocity is = $\frac{1}{0.013} * 0.493^{2/3} 0.267^{1/2}$

v = 24.80 m/s

head loss

$h_f = \frac{0.0019*150*24.80^{2}}{1* 2*9.81}$

$h_f =8.93 m$

pressure difference is

$\frac {p_2- p_1}{\rho g} = 40 - 8.93 = 31.06 m$

$\frac {p_2- p_1}{\rho g} = 31.06 m$

4 0

3 years ago

Which best describes the body in terms of simple machines?

alex41 [277]

Answer:B

Explanation:

5 0

3 years ago

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio

Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

Answer:

$v=23.6m/s$

Explanation:

Velocity $v_c=18.0km/h$

Radius $r=21m$

initial velocity u $u=19=>5.27778$

Generally the equation for Angle is mathematically given by

$\theta=\frac{v_c}{2r}$

$\theta=\frac{18}{2*21}$

$\theta=0.45$

$\theta=25.7831 \textdegree$

Generally

Height of mass

$h=\frac{rsin\theta}{\theta}$

$h=\frac{21sin25.78}{0.45}$

$h=20.3m$

Generally the equation for Work Energy is mathematically given by

$0.5mv_0^2+mgh=0.5mv^2$

Therefore

$v=\sqrt{u^2+2gh}$

$v=\sqrt{=5.27778^2+2*9.81*20.3}$

$v=23.6m/s$

3 0

3 years ago

A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula

makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0

3 years ago

Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati

snow_tiger [21]

Answer:

$D=0.41m$

Explanation:

From the question we are told that:

Discharge rate $V_r=0.35 m3/s$

Distance $d=4km$

Elevation of the pumping station $h_p= 140 m$

Elevation of the Exit point $h_e= 150 m$

Generally the Steady Flow Energy Equation SFEE is mathematically given by

$h_p=h_e+h$

With

$P_1-P_2$

And

$V_1=V-2$

Therefore

$h=140-150$

$h=10$

Generally h is give as

$h=\frac{0.5LV^2}{2gD}$

$h=\frac{8Q^2fL}{\pi^2 gD^5}$

Therefore

$10=\frac{8Q^2fL}{\pi^2 gD^5}$

$D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}$

$D=0.41m$

8 0

3 years ago