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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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Well Bob would need to calculate to net force of someone going down a water slide. Since the person is going down the slide, the person will go faster, depending on their mass/weight and the gravitational pull.
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3 years ago
A eutectic alloy is which of the following (two correct answers): (a) the composition in an alloy system at which the two elemen
NeTakaya

Answer:

Option (c) and option (d)

Explanation:

Eutectic system is one in which a solid and homogeneous mixture of two or more substances resulting in the formation of super lattice is formed which can melt or solidify at a temperature lower than the melting point of any individual metal.

Eutectic alloys are those which have its components mixed in a specific ratio.

It is the composition in an alloy system for which both the  liquidus and solidus temperatures are equal.

Eutectic alloys have the composition in which the melting point of the metal is lower than the other alloy composition.

3 0
3 years ago
At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r
ludmilkaskok [199]

Answer:

a) COP_{R} = 25.014, b) T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

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

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}

COP_{R} = 25.014

b) The respective coefficient of performance is determined:

COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}

COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}

COP_{R} = 5

But:

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

The temperature at hot reservoir is found with some algebraic help:

COP_{R} \cdot (T_{H}-T_{L})=T_{L}

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

T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}}  \right)

T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5}  \right)

T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)

8 0
3 years ago
Read 2 more answers
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Explanation:

The federal highway administration reports nearly 800 work zone fatalities per year.

8 0
3 years ago
A common procedure for measuring the velocity of an air stream involves insertion of an electrically heated wire (called a hot-w
timurjin [86]

Answer:

V = 6.33 m/s

Explanation:

Given:

- The length of the wire L = 0.02 m

- The diameter of the wire D = 0.0005 m

- The calibration expression V = 0.0000625*h^2

- Environment temperature T_inf = 298 K

- Surface temperature T_s = 348 K

- The voltage drop dV = 5 V

- The electric current I = 0.1 A

Find:

- the velocity of Air

Solution:

- Calculate the surface area of the wire:

                             A = pi*D*L

                             A = pi*(0.0005)*(0.02) = 0.00003142 m^2

- The rate of energy in the wire P:

                             P = I*dV = 0.1*5 = 0.5 W

- Apply Newton's Law of Cooling:

                            P = h*A*(T_s - T_inf)

                            h =  P /A*(T_s - T_inf)

Plug in the values:

                             h= 0.5/ 0.00003142*(348 - 298)

                             h = 318.27 W /m^2K

- Using the calibration relationship given, compute the velocity of air:

                             V = 6.25*10^-5 * h^2

                             V = 6.25*10^-5 * (318.27)^2

                             V = 6.33 m/s

5 0
3 years ago
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