A field sample of an unconfined aquifer is packed in a test cylinder. The length and diameter of the cylinder are 50 cm and 6 cm respectively. The field sample is tested for a period of three minutes under a constant head difference of 16.3 cm. As a result, 45.2 cm3 of water is collected at the outlet. Determine the hydraulic conductivity of the aquifer sample.

Answer: new depth will be 3.462m and the water elevation will be 0.462m.

The maximum contraction will be achieved in width 0<w<3

Explanation:detailed calculation and explanation is shown in the image below

Answer

The Impedance Z = 20.982 ohms

The phase angle is Φ= - 70.51°

The current I= 0.477amp

Explanation:

This problem bothers on alternating current, this time an R-L-C circuit

What is a R-L-C circuit?

An RLC circuit is an electrical circuit consisting of a resistor, an inductor, and a capacitor, connected in series or in parallel. The name of the circuit is derived from the letters that are used to denote the constituent components of this circuit.

N/B : Kindly find attached solutions and diagrams for your reference

Answer:

Since the Reynolds number is below 3 x 10⁵ the use of ASTM B152 plate WILL NOT be in complaince with ASME Code for process piping.

Explanation:

Detailed Explanation is given in the attached document.

Answer:

[tex]\Delta p_{2} = 2\cdot \Delta p_{1}[/tex]

Explanation:

The pressure drop is directly proportional to the length of the pipe. Then, the new pressure drop is two time the previous one.

[tex]\Delta p_{2} = 2\cdot \Delta p_{1}[/tex]

Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

[tex]\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}[/tex]

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

[tex]\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m[/tex]

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

[tex]n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times[/tex]

c) By using the equation of the ideal gases you obtain:

[tex]PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules[/tex]

Answer:

The force exerted by the water on one side of the plate is F = 24*pg  

Explanation:

From the given question, the first step to take is to find  he force exerted by the water on one side of the plate.

Solution

Given that:

Let the pressure the  at a depth of y ft be = pgy lb/Pa

the area of the atrip is given as = f(y)*delta(y) = 4/3*(y-2)delta(y)

Then

we combine with the range for y as = y E [2 , 5]

Thus,

F = 4/3*pg * integral from (2 , 5) [y(y-2)] dy

Recall that,

p = water density

g= gravity of acceleration

so,

F = 4/3*pg * integral from (2 , 5) [y^2 - 2y]dy]

F = 4/3*pg * [y^3/3 - y^2] [2 , 5]

F = 4/3*pg * [18]

Finally, F = 24*pg  

Answer:

Horse power = 167.84 hp

Explanation:

Horsepower is calculated using the formula;

P = T * w

See the attached file for the calculation

Answer:

See explaination

Explanation:

The Fourier transform of y(t) = x(t - to) is Y(w) = e- jwto X(w) . Therefore the magnitude spectrum of y(t) is given by

|Y(w)| = |X(w)|

The phase spectrum of y(t) is given by

<Y(w) = -wto + <X(w)

please kindly see attachment for the step by step solution of the given problem.

Answer:

see explaination

Explanation:

The statistical procedure for comparing the 3 groups was the F test with 2 degrees of freedom in the numerator (groups - 1) and 23 df in the denominator (N total - groups)

It is calculated as SSB/2/(SSW/23) = 10.22754

As Fcrit =

from Excel, finv(.05,2,23) = 3.4221, we can reject the null hypothesis of no difference between groups.

SSW = the sum of std i ^ 2 * (n i - 1)

SSB = the sum of ni (mean i - mean)^2

2. From Excel, we get the pvalue from fdist(10.22754,2,23) = .000664

3. For LSD, we calculate (mean 1 - mean 2)/(s * sqrt(1/n1+1/n2))

This is the pooled s = sqrt(SSW/23)

Then, we found t crit from tinv(.05,23) = 2.068

Making use of the chart i made, We found significant differences between std and lac as well as std and veg, but no significant difference between lac and veg.

Answer:

Cv = 0.026 cm²/min

t  = 52.60 min

v% = 41.86 %

tv = 0.1375

t = 8.53 min

v = 53.61 %

Explanation:

given data

height = 2.54 cm

50 % consolidation = 12 min

solution

we get here first Cv value that is express as

Tv = [tex]\frac{Cv\times t}{d^2}[/tex]    .................1

here Tv for 50% is 0.196

put here value and we get

0.196 = [tex]\frac{Cv\times 12}{\frac{2.54}{2}^2}[/tex]  

solve it we get

Cv = 0.026 cm²/min

and

for tv for 90 % consolidation is 0.848

put value in equation 1

0.848 =  [tex]\frac{0.026\times t}{\frac{2.54}{2}^2}[/tex]  

solve it we get t

t  = 52.60 min

and

v% will be here  is

v% = [tex]\frac{0.18}{0.43} \times 100[/tex]  

v% = 41.86 %

and

tv = [tex]\frac{\pi }{4}\times \frac{4}{100}^2[/tex]  

tv = 0.1375

so now put value in equation 1 we get

0.1375 = [tex]\frac{0.026 \times t}{\frac{2.54}{2}^2}[/tex]  

solve it we get

t = 8.53 min

and

now put value of t 14 min in equation 1 will be

tv = [tex]\frac{0.026 \times 14}{\frac{2.54}{2}^2}[/tex]  

t =  0.225 min

and v will be after 14 min

0.0225 = [tex]\frac{\pi }{4}\times \frac{v}{100}^2[/tex]  

v = 53.61 %

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

Answer:

[tex]v_{o} = 22.703\,\frac{m}{s}[/tex] [tex]\left(50.795\,\frac{m}{s}\right)[/tex]

Explanation:

The deceleration of the car on the dry pavement is found by the Newton's Law:

[tex]\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}[/tex]

Where:

[tex]a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g[/tex]

[tex]a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]a_{1} = -3.040\,\frac{m}{s^{2}}[/tex]

Likewise, the deceleration of the car on the unpaved shoulder is:

[tex]a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g[/tex]

[tex]a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]a_{2} = -2.549\,\frac{m}{s^{2}}[/tex]

The speed just before the car entered the unpaved shoulder is:

[tex]v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}[/tex]

[tex]v_{o} = 18.175\,\frac{m}{s}[/tex]

And, the speed just before the pavement skid was begun is:

[tex]v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}[/tex]

[tex]v_{o} = 22.703\,\frac{m}{s}[/tex] [tex]\left(50.795\,\frac{m}{s}\right)[/tex]

The initial speed of the vehicle just before the pavement skid was begun is 5284.65 ft/hr.

Dry pavement friction coefficient Fdry = 0.33

Length of skid marks on dry pavement ddry = 100 ft

Friction coefficient on unpaved shoulder Fshoulder = 0.28

Length of skid marks on unpaved shoulder = 200 ft

First, let's calculate the work done on dry pavement:

Work on dry pavement = Fdry × ddry = [tex]0.33 *100[/tex]

= 33 ft·lbf

Work on unpaved shoulder = Fshoulder × dshoulder

= [tex]0.28 * 200[/tex]

= 56 ft·lbf

Total work done = Work on dry pavement + Work on unpaved shoulder = 33 + 56

= 89 ft·lbf

Assuming the car's mass remains constant, and the final speed is 0, we have:

89 ft·lbf = (1/2)m × (10 mi/hr)²

Convert the final speed to feet per hour:

[tex]10 mi/hr = 10 × 5280/3600 = 5280 ft/hr[/tex]

Now, solve for the initial speed:

v = √((2 × 89 ft·lbf) / m)

v ≈ √((2 × 89) / (6.38 × 10⁻⁶)) ft/hr

v ≈ [tex]\sqrt{(27889055.9}[/tex] ft/hr

v ≈ 5284.65 ft/hr

Answer:

Check the explanation

Explanation:

Kindly check the attached image below for the full step by step explanation to your question.

Answer: −3.46kJ/K

Explanation:

From the question above, we have:

The mass of the block (m) = 45kg

The initial temperature of the block (T1) = 280∘C

The weight of the water (mw) = 100kg

The temperature of water (Tw) = 18∘C

Recall the energy balance equation,

ΔUI = −ΔUw

In this case ΔUI is the internal energy of the iron, while ΔUw is the internal energy of water.

[mcp (T2 − T1)]I = −[mcp (T2 − T1)]w

Here cp is the specific heat at constant pressure.

The specific heat of iron is (cp)I = 0.45kJ/kg⋅K, and the specific heat of water is (cp)w = 4.18kJ/kg⋅K.

Now, we substitute the values in above equation,

[45 × 0.45(T2 − 280)]I = −[100 × 4.18(T2 − 18)]w

[20.25(T2 − 280)] = −[418(T2 − 18)]

20.25T2 − 5,670 = −[418T2 − 7,524]

20.25T2 − 5,670 = −418T2 + 7,524

20.25T2 + 418T2 = 7,524 + 5,670

438.35T2 = 13,194

T2 = 30.1K

Recall, the expression to calculate the total entropy change is given as:

ΔStotal = ΔSI + ΔSw

ΔStotal = [mcpln(T2/T1)]I + [mcpln(T2/T1)]w

Now, we substitute the values in above equation,

ΔStotal = [45 × 0.45ln(297.6/553)]I + [100 × 4.18ln(297.6/291)]w

ΔStotal = −12.55 + 9.09

ΔStotal = −3.46kJ/K

Thus the total entropy change is −3.46kJ/K.

Answer:

Both Technician A, and Technician B are correct

Explanation:

The Traction control are found in those modern automobile, it's a part of the electronic stability control and it becomes active once the automobile get acceleration. It helps the tired of the car not to slip when the car speed up.

It functions by making the car wheel to stop spinning through the reduction of power that is transferred to the wheel i.e application of traction on the wheels of the car. when car is moving with acceleration on a road with with little friction, the Traction is used.

During raining or snow when the road become slippery , In the old cars that doesn't have traction control, the gas pedal is feathered. Which helps to function as traction control

The evidence that indicates that a reaction has occurred include the following:

B. A solid brown product formed.

C. The temperature increased.

D. A gas formed.

E. An explosion occurred.

A chemical reaction is a chemical process that involves the continuous transformation (rearrangement) of the ionic, atomic or molecular structure of a chemical element by breaking down and forming chemical bonds, in order to produce a new chemical compound while new bonds are formed.

This ultimately implies that, a chemical change would give rise to the chemical properties of matter by causing the transformation of one chemical substance into one or more different chemical substances.

Answer:

In place of spiral configuration, we can also use metal wire rings which will provide better strength against circumferential stress. But this spiral configuration will provide strength against circumferential along with axial stress also. If the vascular prosthesis gets elongated along axial direction the corresponding radius of the rings will become short and it will prohibit the expansion of the vascular prosthesis along the circumferential direction.

Explanation:

The orientation is attached below.

Answer:

The pressure and power of fan is 1.77 and 11.18 Hp respectively.

Explanation:

Given:

Discharge [tex]Q_{1} = 30000[/tex] cfm

Pressure difference [tex]\Delta P = 4[/tex] inch

Efficiency [tex]\eta = 50\%[/tex]

(A)

From the formula of fan power,

     [tex]P _{1} = \frac{Q \Delta P}{6356 \eta}[/tex]

     [tex]P_{1} = \frac{30000 \times 4}{6356 \times 0.5}[/tex]

     [tex]P_{1} = 37.76[/tex] Hp

(B)

Fan power and pressure is given by,

We know that pressure difference is proportional to the square of discharge.

    [tex]\frac{\Delta p_{2} }{\Delta P_{1} } = (\frac{Q_{2} }{Q_{1} } ) ^{2}[/tex]

   [tex]\Delta P_{2} = (\frac{20000}{30000} ) \times 4[/tex]

   [tex]\Delta P_{2} = 1.77[/tex]

Fan power proportional to the cube of discharge.

       [tex]\frac{P_{2} }{P^{1} } = (\frac{Q_{2} }{Q_{1} } )^{3}[/tex]

       [tex]P_{2} = \ (\frac{20000}{30000} ) ^{3} \times 37.76[/tex]

       [tex]P_{2} = 11.18[/tex] Hp

Therefore, the pressure and power of fan is 1.77 and 11.18 Hp respectively.

Answer:

Explanation:

By using Bernoulli's Equation:

[tex]\frac{P_1}{P_g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{P_g}+\frac{v_2^2}{2g}+z_2+f\frac{L}{D}\frac{v^2}{2g}[/tex]

where;

[tex]z_1 = z_2 \ and \ v_1 = v_2[/tex]

[tex]P_1 - P_2 = f \frac{L}{D}\frac{1}{2}\rho v^2[/tex]

[tex]P_1-P_2 = \frac{5 \ lb}{in^2}( 144 \frac{in^2}{ft^2})[/tex]

[tex]P_1-P_2 = 720 \frac{lb}{ft^2}[/tex]

[tex]V = \frac{Q}{A} \\ \\ V = \frac{6.684 \ ft^2/s}{\frac{\pi}{4}D^2} \\ \\V = \frac{8.51}{D^2}[/tex]

Density of gasoline [tex]\rho = 1.32 \ slug/ft^3[/tex]

Dynamic Viscosity [tex]\mu[/tex] = [tex]6.5*10^{-6} \frac{lb.s}{ft}[/tex]

[tex]P_1-P_2 = f \frac{L \rho V^2}{2D}[/tex]

[tex]720 = f \frac{L(100)(1.32)}{2D}(\frac{8.51}{D^2})^2[/tex]

D = 1.46 f

[tex]Re, = \frac{\rho VD}{\mu} = \frac{1.32 *\frac{8.51}{D^2} D}{6.5*10^{-6}}[/tex]

[tex]= \frac{1.72*10^6}{D}[/tex]

[tex]\frac{E}{D}= \frac{0.00015}{D}[/tex]

However; the trail and error is as follows;

Assume ; f= 0.02 → D = 0.667ft[tex]\left \{ {{Re=2.576*10^6} \atop {\frac{E}{D}=0.000225}} \right.[/tex]   [tex]\right \{ {{f=0.014} \atop {\neq 2}}[/tex]

f = 0.0145  → D = 0.0428 ft [tex]\left \{ {{Re=4.018*10^6} \atop {\frac{E}{D}=0.00035}} \right.[/tex] [tex]\right \{ {{f=0.015} \atop {\neq 0.0145}}[/tex]

f = 0.0156  → D = 0.43 ft [tex]\left \{ {{Re=4.0*10^6} \atop {\frac{E}{D}=0.000348}} \right.[/tex] [tex]f = 0.0156[/tex]

∴ pipe diameter d = 0.43 ft

Given that:

D = 1 ft

[tex]V = \frac{Q}{A} \\ \\ V = \frac{6.684}{\frac{\pi}{4}(1)^2} \\ \\ V = 8.51 \ ft/s[/tex]

[tex]Re = \frac{\rho \ V \ D}{\mu } \\ \\ Re = \frac{1.32 *8.51 *1 }{6.5*10^{-6}}[/tex]

[tex]Re = 1.72 *10^6[/tex]

[tex]\frac{E}{D} = \frac{0.00015}{1} \\ \\ = 0.00015[/tex]

[tex]f = 0.0136[/tex]

[tex]P_2-P_1 = \frac{fL \rho V^2 }{2D}[/tex]

[tex]P_2-P_1 = \frac{0.036(100)(1.32)(8.51)^2 }{2*1}[/tex]

[tex]P_2-P_1 = 65 \frac {lb}{ft^2}[/tex]  to psi ; we have:

[tex]P_2-P_1 = 0.45 \ psi[/tex]

Answer: 1) tensile

2) 45 degree orientation

3) principal shear stress

Explanation:

An element representing maximum in-plane shear stress with the associated _tensile normal stresses is oriented at an angle of ____45_degree__ from an element representing the ____principal shear____ stresses.

Answer:

[tex]P = 150.335\,kPa[/tex] (Option B)

Explanation:

The absolute pressure of the air-filled tank is:

[tex]P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)[/tex]

[tex]P = 150.335\,kPa[/tex]

Answer:

864 KN

Explanation:

Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.

Please kindly check attachment for the step by step solution of the given problem.

Answer: 112 + 19.27

Explanation:

Super elevation is an inward transverse slope provided through out the length of the horizontal curve which ends up serving as a counteract to the centrifugal force and checks tendency of overturning. It changes from infinite radius to radius of a transition curve.

Super curve elevation (e) = 4%

4/100= 0.04

e= V^2/gR

Make R the subject of the formula.

egR= V^2

R= V^2/eg

V= 45mph

=45 × 0.44704m/s

=20.1168m/s

g (force due to gravity) =9.81

Therefore,

R= (20.1168)^2/9.81 × 0.04

= 1031.31m

Tangent Length( T) = PI - PC

Tangent Length= 10875 - 10500

=375m

T= R Tan(I/2)

375= 1031.31 × Tan(I/2)

I= 39.96

Also,

L= πRI/180

= 719.27m

Station PT= Stat PC+ L

10500 + 719.27

=11219.27

=112 + 19.27

Answer:

Maximum Normal Stress σ = 8.16 Ksi

Maximum Shearing Stress τ = 4.08 Ksi

Explanation:

Outer diameter of spherical container D = 17 ft

Convert feet to inches D = 17 x 12 in = 204 inches

Wall thickness t = 0.375 in

Internal Pressure P = 60 Psi

Maximum Normal Stress σ = PD / 4t

σ = PD / 4t

σ = (60 psi x 204 in) / (4 x 0.375 in)

σ = 12,240 / 1.5

σ = 8,160 P/in

σ = 8.16 Ksi

Maximum Shearing Stress τ = PD / 8t

τ = PD / 8t

τ = (60 psi x 204 in) / (8 x 0.375 in)

τ = 12,240 / 3

τ = 4,080 P/in

τ = 4.08 Ksi

Answer:

Explanation:

Write the equation for Reynolds number as follows:

Re = VL/v

For dynamic similarity,

(VL/v)m + (VL/v)p…… (1)

Since, the model and prototype are in same medium, the kinematic viscosity remains same.

From equation (1), we can write

(VL)m = (VL)p

Here, L represents length, and V is the velocity.

Re-write the equation as follows:

Vm = Lp/Lm x Vp

Substitute 1/8 for Lp/Lm and 1.5m/s for Vp .

Vm = 1/8 x 1.5

Vm = 0.1875m/s

Therefore, the wind tunnel air speed is 0.1875m/s.

Answer:

Explanation:

check the attached files for solution.

Answer: the metal will experience a strain of approximately 2.037Mpa

This strain is lesser than if the force was applied at room temperature.

This will reduce internal stress and increase some mechanical properties of the aluminum such as mechanical hardening.

Explanation:

Detailed explanation and calculation and comparison with equivalent tensile stretching at ordinary room temperature is compared.

Answer:

Bending stress = 32.29ksi ∠ 33.0 ksi

Explanation:

CHECK BELOW FOR THE ANSWER IN THE FILE ATTACHED.

a. This is a counter flow heat exchanger.

b. The effectiveness of  heat exchanger is 0.615

The  effectiveness of  heat exchanger is defined as ratio of actual heat transfer to the maximum possible heat transfer.

Given that the cold oil enters at 20°C and leaves at 55°C, while the hot oil enters at 80°C and leaves at 45°C.

               [tex]Effectiveness=\frac{55-15}{80-15}=\frac{40}{65}=0.615[/tex]

Learn more:

https://brainly.com/question/17029235

Answer:

a. 0.02639

b. 1.98H

Explanation:

Please see attachment