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Masja [62]
3 years ago
8

The elastic settlement of an isolated single pile under a working load similar to that of piles in the group it represents, is p

redicted to be 0.25 inches. What is the expected settlement for the pile group given the following information?
1.Group: 16 piles in a 4x4 group
2.Pile Diameter: 12 inches
3.Pile Center to Center Spacing: 3 feet

Engineering
1 answer:
Ganezh [65]3 years ago
3 0

Answer:

The expected settlement for the pile group using the given information is 19.92mm or 0.79 inch

Explanation:

In this question, we are asked to calculate the expected settlement for the pole group given some information.

Please check attachment for complete solution and step by step explanation

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Determine if the following errors are systematic or random. Justify your response. (a) Effect of temperature on the circuitry of
k0ka [10]

Answer:

a) temperature: random error

b) parallax: systematic error

c) using incorrect value: systematic error

Explanation:

Systematic errors are associated with faulty calibration or reading of the equipments used and they could be avoided refining your method.

4 0
3 years ago
A fully braced structural member in a building is subjected to several different loads, including roof loads of D = 5 k and L_r
allochka39001 [22]

Answer: 37.4K

Explanation:

See attachment

7 0
3 years ago
Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure i
Brilliant_brown [7]

Answer:

\dot Q_{in} = 372.239\,MW

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

w_{in} + h_{in}- h_{out} = 0

h_{out} = w_{in}+h_{in}

h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}

h_{out} = 203.81\,\frac{kJ}{kg}

The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

h_{out} = 2685.4\,\frac{kJ}{kg}

The rate of heat transfer in the boiler is:

\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)

\dot Q_{in} = 372.239\,MW

3 0
3 years ago
Read 2 more answers
The given family of functions is the general solution of the differential equation on the indicated interval.Find a member of th
Alja [10]

Answer:

Explanation:

y'''+y=0---(i)

General solution

y=c_1e^o^x+c_2\cos x +c_3 \sin x\\\\\Rightarrow y=c_1+c_2 \cos x+c_3 \sin x---(ii)\\\\y(\pi)=0\\\\\Rightarrow 0=c_1+c_2\cos (\pi)+c_3\sin (\pi)\\\\\Rightarrow c_1-c_2=0\\\\c_1=c_2---(iii)

y'=-c_2\cos x+c_3\cosx\\\\y'(\pi)=2\\\\\Rightarrow2=-c_2\sin(\pi)+c_3\cos(\pi)\\\\\Rightarrow-c_2(0)+c_3(-1)=2\\\\\Rightarrow c_3=-2\\\\y''-c_2\cos x -c_3\sin x\\\\y''(\pi)=-1\\\\\Rightarrow-1=-c_2 \cos (\pi)=c_3\sin(\pi)\\\\\Rightarrow-1=c_2-0\\\\\Rightarrow c_2=-1

in equation (iii)

c_1=c_2=-1

Therefore,

\large\boxed{y=-1-\cos x-2\sin x}

5 0
3 years ago
A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a
statuscvo [17]

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre =  183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

3 0
3 years ago
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