A site is underlain by two layers of normally consolidated clayey sand. The unit weights of the top layer are 19 and 21 kN/m2 and the layer is 6 meters thick. The unit weights of the bottom layer are 20 and 22 kN/m, and the layer is 8 meters thick. Below the bottom layer lies bedrock. The water table is located 2 meters below the ground surface. The top layer soil has a friction angle of 38 degrees, a cohesion intercept of 20 kPa, and an undrained strength of 160 kPa. The bottom layer soil has a friction angle of 33 degrees, a cohesion intercept of 10 kPa, and an undrained strength of 120 kPa. Point A is located 9 meters underground. All layers have a lateral stress ratio of 0.5.

Required:
a. Draw the profile neatly, add dimensions, and draw a tree on the ground surface.
b. Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A.
c. Draw the Mohr Circle to determine the effective stress that acts at point A on a plane inclined 10 degrees counter-clockwise from the horizontal plane. Make sure the Mohr Circle is a well-drawn circle

Answer:

Hello your question is incomplete attached below is the missing part of the question

answer ;

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

Explanation:

Determine voltage drop from top terminal to bottom terminal ( Vab ) in the right hand branch and Vcd in left hand branch

40v and 50mA are in series hence;  Ix = 50mA

also Vcd = 30V

CD is parallel to AB hence; Vcd = Vab = 30 V

Vab = ∝*Ix + 60 v

 30v  = ∝ ( 50mA ) + 60

therefore ∝ = -600

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

Answer:

50421.6 m³

Explanation:

The river has an average rate of water flow of 59.6 m³/s.

Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:

Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s

The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken

time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds

The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³

Answer:

- Vernier thrust is 5.37 kN

- mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s

Explanation:  

Given that;

For individual thrust chambers (vacuum);

Fc = 73.14 kN ,  Cc = 3279 m/sec

For Overall engine with Vernier (vacuum);

Foa = 297.93 kN = ,  Coa = 3197 m/sec.

- determine the Vernier thrust

Vernier thrust Fv =  Foa - ( 4 × Fc )

Vernier thrust Fv  = 297.93 - ( 4 × 73.14)

Vernier thrust Fv  = 297.93 - 292.56

Vernier thrust Fv  = 5.37 kN

Therefore, Vernier thrust is 5.37 kN

-

Vernier mass flow rate;

we know that

[tex]Co_{a}[/tex] = Fc + Fv  / mc + mv

mv = Foa/Coa - Fc/Cc

we convert kilonewton to kilograms

1 kn = 102 kg

Fc = 73.14 kN = 73.14 × 102 = 7460.28 kg

Foa = 297.93 kN = 297.93 × 102 = 30388.86 kg

we substitute

mv = (30388.86 / 3197) - (( 4 × 7460.28) / 3279)

mv = 9.5054 - 9.1006

mv = 0.4048 kg/s

Therefore, mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s

Answer:

The answers are below

Explanation:

a. 180 in^3 to L

1 in³ = 0.0164L

180 in³ = [tex]180\ in^3*\frac{0.0164\ L}{1\ in^3}= 2.95\ L[/tex]

b. 750 ft-lbf to kJ

1 ft-lbf = 0.00136 kJ

750 ft-lbf  = [tex]750\ ft-lbf *\frac{0.00136\ kJ}{1\ ft-lbf} =1.02\ kJ[/tex]

c. 75.0 hp to kW

1 hp = 0.746 kW

75 hp = [tex]75\ hp*\frac{0.746\ kW}{1\ hp}=55.95\ kW[/tex]

d. 2500.0 lb/h to kg/s

1 lb/h = 0.000126 kg/s

2500.0 lb/h = [tex]2500.0\ lb/h*\frac{0.000126\ kg/s}{1\ lb/h} =0.315\ kg/s[/tex]

e. 120 psia to kPa

1 psia = 6.89 kPa

120 psia = [tex]120\ psia*\frac{6.89\ kPa}{1\ psia} =826.8\ kPa[/tex]

f. 120 psig to kPa

1 psig = 6.89 kPa

120 psig = [tex]120\ psia*\frac{6.89\ kPa}{1\ psig} =826.8\ kPa[/tex]

g. 300 ft/min to m/s

1 ft/min = 0.005 m/s

300 ft/min = [tex]300\ ft/min*\frac{0.005\ m/s}{1\ ft/min} = 1.5\ m/s\\[/tex]

h. 125 km/h to miles/h

1 km/h = 0.62 mph

125 km/h = [tex]125\ km/h*\frac{0.62\ mph}{1\ km/h} =77.5\ mph[/tex]

i) 6000 N to Ibf

1 N = 0.2248 lbf

6000 N = [tex]6000\ N*\frac{ 0.2248\ lbf}{1\ N}=1348.8\ N[/tex]

j. 6000 N to ton

1 N =  0.000102 Ton-force

6000 N = [tex]6000\ N*\frac{ 0.000102\ Ton-force}{1\ N}=0.612\ N[/tex]

Answer:

a) attached below

b) The worst case in tension is case ( 9-7 ) b which is

  = -65k

  The worst case in compression is case ( 9-5 ) a which is

 = 670 k  

Explanation:

Given data :

D = 150k , L = 280k , Lr = 40k , s = 50k , w = ± 120k

E = ± 200k

attached below is a detailed solution to the given problem ( problem 1 )

A)  attached below

b) The worst case in tension is case ( 9-7 ) b which is

  = -65k

  The worst case in compression is case ( 9-5 ) a which is

 = 670 k  

Answer:

Attached below

Explanation:

PWM signal source has 1 KHz base frequency

Analog filter : with time constant = 0.01 s

low pass transfer function = [tex]\frac{1}{0.01s + 1 }[/tex]

PWM duty cycle is a constant block

Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively

Answer: Why is even here then.

Explanation:

Answer:

685.38 MJ

Explanation:

Given that:

mass = 10 tons = 1.0 × 10 ⁴ kg

diameter D = 10 m

radius R = 5 m

speed N = 1000 rpm

Using the formula for K.E = [tex]\dfrac{1}{2}I \omega^2[/tex] to calculate the energy stored

where;

[tex]= \dfrac{2 \pi \times N}{60}[/tex]

[tex]= \dfrac{2 \pi \times 1000}{60}[/tex]

= 104.719 rad/s

Hence, the energy stored is;

[tex]= \dfrac{1}{2}\times (\dfrac{MR^2}{2}) \times \omega^2[/tex]

[tex]= \dfrac{1}{2}\times (\dfrac{10^4\times 5^2}{2}) \times 104.719^2[/tex]

= 685379310.1

= 685.38 MJ

Answer:

a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b) length of the specimen is 66.97 mm

Explanation:

Given the data in the question;

a) Determine the maximum stress that can be applied without plastic deformation

when know that; maximum stress σ[tex]_{max}[/tex]  = F / A

where F is the force in the rod ( 39872 N )

A is the cross-sectional area of the rod ( 100 mm² )

so we substitute;

σ[tex]_{max}[/tex]  = 39872 N / 100 mm²

σ[tex]_{max}[/tex]  = 398.72 N/mm²

Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b)  

strain in the members can be calculated using the expression

ε = σ / E

where σ is the stress in the rod

E is the module of elasticity (  110 GPa = 110000 N/mm² )

(Sl-L) / L = σ/E

where Sl-L is the change in length of the member

L is the original length of the specimen

so we substitute

(67.21 - L) / L = 398.72 / 110000

110000( 67.21 - L) = 398.72L

7393100 - 110000L = 398.72L

7393100 = 398.72L+ 110000L

7393100 = 110398.72L

L = 7393100 / 110398.72

L = 66.97 mm

Therefore; length of the specimen is 66.97 mm

Answer:

30 mm is the minimum thickness that must be applied.

Explanation:

Given the data in the question;

Using Fourier's equation. the heat rate is  

q = kA(ΔT/Δx)

where

A is the surface area, we must consider all surfaces through which the heat can dissipate through

i.e 2×2 for one wall gives you 4m²,

there are 5 walls, so we will  have 20m² for surface area.

k is thermal conductivity of the styrofoam ( 0.030 W/m K)    

q is the heat loss (500 W  )

ΔT is the Temperature difference ( 35 - 10) = 25°C

Δx  = ?

So we substitute

500 = (0.030)(20)(25/Δx)

500 = 0.6 (25/Δx)

500 = 15 / Δx

Δx = 15 / 500

Δx = 0.03 m = 30 mm

Therefore, 30 mm is the minimum thickness that must be applied.

Answer:

no it has to be removed

Explanation:

It is completely inappropriate to mention that the drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been removed. Therefore, the statement given above is false.

Drum brakes can be referred to or considered as the types of brakes that are useful in application of brakes to an object, such as wheels, in motion. To understand better, it can be stated that the system of braking under drum brakes is completely in contrast to that of disc brakes.

Drum brakes have a hydraulic pressure, which means that if the condition of lining is to be checked, removal of drums becomes essential. If the drums are not removed, correction or alignment of wheels cannot be performed.

Therefore, the significance regarding drum brakes has been aforementioned, and the statement given above with respect to their removal also holds false.

Learn more about drum brakes here:

https://brainly.com/question/14937026

#SPJ2

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W