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Naddika [18.5K]
3 years ago
15

A shrinkage limit test is performed on a soil. The initial mass and volume of the soil are: V1=20.2cm^3 , while the final mass a

nd volume are M2=24g and V2=14.3cm^3 . Note that in the initial state the soil is saturated, whereas in the final state the soil is completely dry.
Calculate:
a. the shrinkage limit SL of the soil.
b. the void ratio at the SL.
c. Gs of the soil solids.
d. the initial void ratio.
Engineering
1 answer:
love history [14]3 years ago
3 0
C. Gs of the soil solids
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): drivers must slow down from 60 to 40 mi/hr to negotiate a severe curve. A warning sign is visible for a distance of 120 ft. H
Yakvenalex [24]

Answer:

Explanation:

Total Stopping distance is the sum of the reaction distance and the braking distance, such that ;

d=dr + db =1.47St +[ ( Si 2- Sf2 ) / 30(F/0.01G) ]

In this case, the reaction time, t, is the AASHTO standard, or 2. 5 s. The friction factor, F, is based upon the standard AASHTO deceleration rate of 11.2 ft/s2 (F = 11.2/32.2 = 0.348) and the speed is given as 40 mph.

d = 1.47x60x2.5+ [ (60^2-40^2) / 30x 0.348 ]

d = 220.5 + ( (3600-1600) / 10.44 )

d = 220.5 + 191.57

d = 412.07ft.

Since the sign can be seen clearly at 120 ft.

Then the position of the sign should be,

= 412.07 - 120

= 292.07 ft

8 0
3 years ago
g A food department is kept at -12oC by a refrigerator in an environment at 30oC. The total heat gain to the food department is
boyakko [2]

Answer:

a) \dot W = 0.417\,kW, b) COP_{R} = 2.198, c) Irreversible.

Explanation:

a) The power input required by the refrigerator is:

\dot W = \dot Q_{H} - \dot Q_{L}

\dot W = \left(4800\,\frac{kJ}{h} - 3300\,\frac{kJ}{h}\right)\cdot \left(\frac{1}{3600} \,\frac{h}{s} \right)

\dot W = 0.417\,kW

b) The Coefficient of Performance of the refrigerator is:

COP_{R} = \frac{\dot Q_{L}}{\dot W}

COP_{R} = \frac{3300\,\frac{kJ}{h} }{(0.417\,kW)\cdot \left(3600\,\frac{s}{h} \right)}

COP_{R} = 2.198

c) The maximum ideal Coefficient of Performance of the refrigeration is given by the inverse Carnot's Cycle:

COP_{R,ideal} = \frac{T_{L}}{T_{H}-T_{L}}

COP_{R,ideal} = \frac{261.15\,K}{303.15\,K - 261.15\,K}

COP_{R,ideal} = 6.218

The refrigeration cycle is irreversible, as COP_{R} < COP_{R,ideal}.

3 0
3 years ago
If x &lt; 5 and x &gt;c, give a value of c such that there
Arlecino [84]

we have  

x<5

x>c

we know that

The solution is the intersection of both solution sets of the given inequalities.  

The solutions of the compound inequality must be solutions of both inequalities.  

The value of c could be 5 or any number greater than 5, such that there are no solutions to the compound inequality

Because

A number cannot be both less than 5 and greater than 5 at the same time

therefore

the answer is

for c_> there are no solutions to the compound inequality

7 0
3 years ago
The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The b
Wewaii [24]

Answer:

a) P ≥ 22.164 Kips

b) Q = 5.4 Kips

Explanation:

GIven

W = 18 Kips

μ₁ = 0.30

μ₂ = 0.60

a) P = ?

We get F₁  and F₂ as follows:

F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips

F₂ = μ₂*Nef = 0.6*Nef

Then, we apply

∑Fy = 0   (+↑)

Nef*Cos 12º -  F₂*Sin 12º = W

⇒   Nef*Cos 12º -  (0.6*Nef)*Sin 12º = 18

⇒   Nef = 21.09 Kips

Wedge moves if

P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º

⇒  P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º

⇒  P ≥ 22.164 Kips

b) For the static equilibrium of base plate

Q = F₁ = 5.4 Kips

We can see the pic shown in order to understand the question.

7 0
3 years ago
Read 2 more answers
A police officer in a patrol car parked in a 70 km/h speed zone observes a passing automobile traveling at a slow, constant spee
Ludmilka [50]

Answer:

S = 0.5 km

velocity of motorist = 42.857 km/h

Explanation:

given data

speed  = 70 km/h

accelerates uniformly = 90 km/h

time = 8 s

overtakes motorist =  42 s

solution

we know  initial velocity u1 of police = 0

final velocity u2 = 90 km/h = 25 mps

we apply here equation of motion

u2 = u1 + at  

so acceleration a will be

a = \frac{25-0}{8}

a = 3.125  m/s²

so

distance will be

S1 = 0.5 × a × t²

S1 = 100 m = 0.1 km

and

S2 = u2 ×  t

S2 = 25  × 16

S2 = 400 m = 0.4 km  

so total distance travel by police

S = S1 + S2

S = 0.1 + 0.4

S = 0.5 km

and

when motorist travel with  uniform velocity

than total time = 42 s

so velocity of motorist will be

velocity of motorist = \frac{S}{t}

velocity of motorist =  \frac{500}{42}  

velocity of motorist = 42.857 km/h

3 0
3 years ago
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