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3 years ago

15

Water flows down a rectangular channel that is 1.2 m wide and 1 m deep. The flow rate is 0.95 m/s. Estimate the Froude number of

the flow.

1 answer:

olga_2 [115]3 years ago

8 0

Answer:

The Froude number of the flow in the channel is 0.253

Explanation:

We know that Froude number for a rectangular channel is given by

$F_{r}=\frac{v}{\sqrt{gy}}$

where,

v = is the velocity of flow

g = is the acceleration due to gravity

y = depth of flow

We know that velocity is calculated from the flow rate as

$v=\frac{Q}{Area}\\\\\therefore v=\frac{0.95}{1.2\times 1}=0.792m/s$

Applying the given values in the equation we obtain Froude number as follows $Hence\\\\F_{r}=\frac{0.792}{\sqrt{9.81\times 1}}=0.253$

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The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

bixtya [17]

Answer:

h = 287.1 m

Explanation:

the density of mercury \rho =13570 kg/m3

the atmospheric pressure at the top of the building is

$p_t = \rho gh = 13570*908*0.73 = 97.08 kPa$

the atmospheric pressure at bottom

$p_b = \rho gh = 13570*908*0.75 = 100.4 kPa$

$\frac{w_{air}}{A} =p_b -p_t$

we have also

$(\rho gh)_{air} = p_b - p_t$

1.18*9.81*h = (100.4 -97.08)*10^3

h = 287.1 m

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3 years ago

To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws

ehidna [41]

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

$E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\$

E=52000Hp.h

$E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\$

E=38724920Wh

$E=52000Hph(\frac{1977378.4 ft lb}{1Hph}$

E=1.028x10^11 ftlb

3 0

3 years ago

I don’t get this it’s hella hard

qwelly [4]

Answer:

V₂ = 20 V

Vt = 20 V

V₁ = 20 V

V₃ = 20 V

I₁ = 10 mA

I₃ = 3.33 mA

It = 18.33 mA

Rt = 1090.91 Ω

Pt = 0.367 W

P₁ = 0.2 W

P₂ = 0.1 W

P₃ = 0.067 W

Explanation:

Part of the picture is cut off. I assume there is a voltage source Vt there?

First, use Ohm's law to find V₂.

V = IR

V₂ = (0.005 A) (4000 Ω)

V₂ = 20 V

R₁ and R₃ are in parallel with R₂ and the voltage source Vt. That means V₁ = V₂ = V₃ = Vt.

V₁ = 20 V

V₃ = 20 V

Vt = 20 V

Now we can use Ohm's law again to find I₁ and I₃.

V = IR

I = V/R

I₁ = (20 V) / (2000 Ω)

I₁ = 0.01 A = 10 mA

I₃ = (20 V) / (6000 Ω)

I₃ = 0.00333 A = 3.33 mA

The current It passing through Vt is the sum of the currents in each branch.

It = I₁ + I₂ + I₃

It = 10 mA + 5 mA + 3.33 mA

It = 18.33 mA

The total resistance is the resistance of the parallel resistors:

1/Rt = 1/R₁ + 1/R₂ + 1/R₃

1/Rt = 1/2000 + 1/4000 + 1/6000

Rt = 1090.91 Ω

Finally, the power is simply each voltage times the corresponding current.

P = IV

Pt = (0.01833 A) (20 V)

Pt = 0.367 W

P₁ = (0.010 A) (20 V)

P₁ = 0.2 W

P₂ = (0.005 A) (20 V)

P₂ = 0.1 W

P₃ = (0.00333 A) (20 V)

P₃ = 0.067 W

7 0

3 years ago

How to write the name (ELAF ADNAN) by using G-codes in milling machine ?? Please help

mel-nik [20]

Answer:

Simple G Code Example Mill - This is a very simple G code example for beginner level ... Run the program on your cnc machine (Safety first, keep a professional around). ... This G code program example don't use Tool radius compensation ...

Explanation:

3 0

3 years ago

A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6

amm1812

1) $a_x=4.287+2.772x\\a_y=-5.579+2.772y$

2) 8.418

Explanation:

1)

The two components of the velocity field in x and y for the field in this problem are:

$u=1.85+2.05x+0.656y$

$v=0.754-2.18x-2.05y$

The x-component and y-component of the acceleration field can be found using the following equations:

$a_x=\frac{du}{dt}+u\frac{du}{dx}+v\frac{du}{dy}$

$a_y=\frac{dv}{dt}+u\frac{dv}{dx}+v\frac{dv}{dy}$

The derivatives in this problem are:

$\frac{du}{dt}=0$

$\frac{dv}{dt}=0$

$\frac{du}{dx}=2.05$

$\frac{du}{dy}=0.656$

$\frac{dv}{dx}=-2.18$

$\frac{dv}{dy}=-2.05$

Substituting, we find:

$a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x$

And

$a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y$

2)

In this part of the problem, we want to find the acceleration at the point

(x,y) = (-1,5)

So we have

x = -1

y = 5

First of all, we substitute these values of x and y into the expression for the components of the acceleration field:

$a_x=4.287+2.772x\\a_y=-5.579+2.772y$

And so we find:

$a_x=4.287+2.772(-1)=1.515\\a_y=-5.579+2.772(5)=8.281$

And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:

$a=\sqrt{a_x^2+a_y^2}=\sqrt{1.515^2+8.281^2}=8.418$

4 0

3 years ago