When will stemuless checks come I was told not to file my taxes because I get a pension check every month direct deposit money into my bank account is this true I don't have to do my taxes because of this

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:

1.5258m

Explanation:

Please see attachment

Answer:

length of cylinder can not calculated

Explanation:

given data

tensile stress = 10,000 psi

Original length = 1

Original cross-sectional area = 0.1 in²

Yield strength, σy = 9 ksi

Young’s modulus, E = 1000 ksi

solution

we can see that here that applied stress is greater than yield stress of material  that is express

1000 ksi  >  9 ksi

so here hooks law and strain relation is not working

so length of cylinder can not calculated

as stress applied 10000 psi

Under 10,000 psi tensile stress, the metal cylinder elongates by 0.01 inches, resulting in a final length of 1.01 inches.

To calculate the elongation (change in length) of the metal cylinder under tensile stress, we can use Hooke's Law, which states that the elongation [tex](\( \Delta L \))[/tex] is directly proportional to the applied tensile stress [tex](\( \sigma \))[/tex] and the original length [tex](\( L_0 \))[/tex], and inversely proportional to the Young's modulus [tex](\( E \))[/tex]:

[tex]\[ \Delta L = \frac{\sigma \cdot L_0}{E} \][/tex]

Given:

- Original length [tex](\( L_0 \))[/tex] = 1 in

- Applied tensile stress [tex](\( \sigma \))[/tex] = 10,000 psi

- Young's modulus [tex](\( E \))[/tex] = 1000 ksi = 1,000,000 psi

Substitute the values into the formula:

[tex]\[ \Delta L = \frac{10,000 \times 1}{1,000,000} \][/tex]

[tex]\[ \Delta L = \frac{10,000}{1,000,000} \][/tex]

[tex]\[ \Delta L = 0.01 \, \text{in} \][/tex]

So, the elongation of the metal cylinder under the given tensile stress is 0.01 inches.

To find the final length, we add the elongation to the original length:

[tex]\[ \text{Final length} = \text{Original length} + \Delta L \][/tex]

[tex]\[ \text{Final length} = 1 + 0.01 \][/tex]

[tex]\[ \text{Final length} = 1.01 \, \text{in} \][/tex]

Therefore, the final length of the metal cylinder under the given tensile stress is 1.01 inches.

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:

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:

Explanation:

Please kindly go through the attached file for a step by step approach to the solution of this problem

Question:

The question is not complete. The question to answer was not added. See below the possible question and the answer.

a. How many blocks are in the cache with this new arrangement?

b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.

C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?

Answer:

(a) Number of blocks =  512 blocks

(b) Tag is 18

(c)  Total size of the cache = 8388608 bytes

Explanation:

a .

block size = 32 bytes

cache size = 16384 bytes

No.of blocks = 16384 / 32

No.pf blocks = 512 blocks

b.

Total address size = 32 bits

Address bits = Tag + Line index +block offset

Block Size = 32 bytes.

So block size = 25 bytes.

Hence Offset is 5

No . of Cache blocks = 512 blocks = 29 blocks

Hence line offset is 9

We know that Address bits = Tag + Line index +block offset

So , 32 =tag+9+5

tag = 32-(9+5)

So Tag is 18

c.

Data bits = 32 bits

Tag=18 bits

Valid bit is 1 bit

so Total cache size = 25+218+20

                                  = 223

                                  =8388608 bytes

Answer:

i) The torque required to raise the load is 15.85 N*m

ii) The torque required to lower the load is 6.91 N*m

iii) The minimum coefficient of friction is -0.016

Explanation:

Given:

dm = mean diameter = 0.03 m

p = pitch = 0.004 m

n = number of starts = 1

The lead is:

L = n * p = 1 * 0.004 = 0.004 m

F = load = 7000 N

dc = collar diameter = 0.035 m

u = 0.05

i) The helix angle is:

[tex]tan\alpha =\frac{L}{\pi *d_{m} } =\frac{0.004}{\pi *0.03} \\\alpha =2.43[/tex]

The torque is:

[tex]T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm[/tex]

ii) The torque to lowering the load is:

[tex]T=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03-0.004}{\pi *0.03+0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=6.91Nm[/tex]

iii)

[tex]T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\ 0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}[/tex]

Clearing u:

u = -0.016

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:

theoretical fracture strength  = 16919.98 MPa

Explanation:

given data

Length (L) = 0.28 mm = 0.28 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa

solution

we get here theoretical fracture strength s that is express as

theoretical fracture strength  =   [tex]s_{0} \times \sqrt{\frac{L}{r} }[/tex]   .............................1

put here value and we get

theoretical fracture strength  =    [tex]1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }[/tex]  

theoretical fracture strength  =  [tex]16919.98 \times 10^6[/tex]  

theoretical fracture strength  = 16919.98 MPa

Answer:

Horse power = 167.84 hp

Explanation:

Horsepower is calculated using the formula;

P = T * w

See the attached file for the calculation

Answer:

R = 7.854 x 10⁵ anpert turns / Wb

Relative permeability = 405.3

Explanation:

Detailed explanation is given in the attached document.

Answer: 85398.16, 405.28473.

Explanation:

We are given that the number of turns on the core is 500 2 , the cross sectional area is :

A=5cm^{2}({1meter}/{100cm})^{2} =0.0005meters^{2}

and the length of the core is l=20cm ({1meter}/{100cm})= 0.2meters .

In this solution, we are meant to neglect the resistance of the coil , and the current through the coil is I=1amPrms when the voltage applied across it is:

V=120voltsrms at f=60Hz. From this, We can calculate the inductance(L) whiof the coil (a coil have an inductance value of one Henry when an electromotive force of one volt is induced in the coil were the current flowing through the said coil changes at a rate of one ampere/second).

The natural frequency of the applied voltage is:

ω =2π f=2π(60)=120π{radians}/{second} .

The inductive reactance of the coil is equal to X=ω L=120π L . We then know that current is :

I=V/X

=I =20/120πL

L=120/120π

=1/π henries .

For reluctance (R) (which is a unit measuring the opposition to the flow of magnetic flux within magnetic materials and is analogous to resistance in electrical circuits). Looking at the relationship between inductance and reluctance . You will note that it is :

L=n^2/R .

We can use this relationship to find reluctance for our closed iron core coil :

L=1/π = 500^2/R

R=500^2π =785398.16{amps}/{volt-seconds}}

We can therefore use the other equation for reluctance.

R=1/μ A= 1/μA or μRA

To calculate the relative permeability of the core :

R=500^2π

=0.2/(4π ×10^-7)/ μR(0.0005)

μR=0.2/(4π ×10^-7)/ 500^2π(0.0005)

= 405.28473

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:

(a) BOD removal efficiency = 89.51%

(b) Aeration tank volume = 4732m³

(c) MLSS = 1706.669 mg/L

(d) Secondary clarifier surface area 6250ft²

(e) Air requirement =  12930.284 lb

Explanation:

See the attached file for explanation.  This is continuation from page 3 of the attached file.

Air required = 172403.792 ft3

since, density of air = 0.075lb/ft3

air required = 0.075*172403.792 lb

                    = 12930.284 lb

Answer:

Activity sequence model = A1B0G1A0B0P3F16A1B0P1A0

Tn= 10(1 + 1 + 3 + 16 + 1 + 1) = 10(23) = 230 TMU (8.3 sec)

Answer:

Explanation:

first convert difference equation to transfer function form,

apply laplase transform to difference equation

s2Y(s) + 6 * s * y(s) + 2 * y(s) = s * X(s) + 4 * X(s)

(s2 + 6s+2) * y(s) = (s+4)*X(s)Y s)

Lets write below code in matlab command prompt

lets use lodspace to create values from 10^-1 to 10^5 and use freqs to plot frequency response of above system with frequency w

>> n=[1 4];

>> d=[1 6 2];

>> w = logspace(-1,5);

>> freqs(n,d,w)

Attached is the written solution and the MATLAB diagram

Answer:

the restoring force is = 3/4NKT

Explanation:

check the attached files for answer.

Answer:

435.032 kj

Explanation:

We can describe Heat transfer as a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems.

Please refer to the attached file for the detailed step by step solution of the given problem

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

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:

Check the explanation

Explanation:

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

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:

60 rpm

Explanation:

At t = 0,

Angular speed = 0

At t = 10 sec

Angular speed = 20/10 = 2 rev/s

Average speed = (2 - 0)/2 = 1 rev/s

= 1 x 60 = 60 rpm

Answer:

a. 0.02639

b. 1.98H

Explanation:

Please see attachment

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:

See explaination

Explanation:

We can say that that the The Darcy Friction factor or Equation is a theoretical equation that predicts the frictional energy loss in a pipe based on the velocity of the fluid and the resistance due to friction. It is used almost exclusively to calculate head loss due to friction in turbulent flow.

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

Answer:

# list of 10 zip codes assigned to zip

zips = ["12789", "54012", "54481", "54982", "60007", "60103", "60187", "60188", "71244", "90210"]

# user is prompt to enter zip code and assigned to user_zip

user_zip = input("Enter your zip code: ")

# if else statement to check if user input is available for delivery

# if statement check if user zip is in zip, if it is, it display

# delivery is okay to specified zip

if user_zip in zips:

   print("Delivery to {} ok.".format(user_zip))

# else it display no delivery to such zip code

else:

   print("Sorry - no delivery to {}.".format(user_zip))

Explanation:

The question doesn't specify programming language to use. Since no programming language was stated, the problem was solved using Python3. List structure is the equivalent of array in Python.

Assumption was also made on the array holding 10 zip codes of areas to which the company make deliveries.

The program first initialized a list of 10 zip codes and assigned it to zips. Then it prompt the user to enter a zip code which is assigned to user_zip.

Then if-else statement is used to check if user inputted zip is available with the zips variable.

If it is available, "Delivery ok" is displayed to the user else "no delivery" is displayed to the user.

Answer:

0.09cm/sec

Explanation:

We are going to describe Hydraulic conductivity as a measure of the ease with which water flows through sediments, determining renewal rates of water, dissolved gases, and nutrients.

See attachment for a detailed solution.

The hydraulic conductivity of the given aquifer sample packed in a test cylinder is; 0.027 cm/s

We are given;

Head Loss; H = 16.3 cm

Length of cylinder; L = 50 cm

Diameter of cylinder; d = 6 cm

radius; r = 6/2 = 3 cm

time; t = 3 minutes = 180 s

Volume; V = 45.2 cm³

Formula for the hydraulic gradient is;

i = H/L

i = 16.3/50

i = 0.326

Formula for volumetric rate of flow is;

Q = V/t

Q = 45.2/180

Q = 0.25 cm³/s

Formula for area is;

A = πd²/4

A = π × 6²/4

A = 9π

Formula for hydraulic conductivity is;

K = Q/(A * i)

K = 0.25/(9π * 0.326)

K = 0.027 cm/s

Read more about Aquifers at; https://brainly.com/question/1052965

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