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4 years ago

Stone has been used as a building material since ancient times because of its compressive strength, which is the __________.

Engineering

1 answer:

Doss [256]4 years ago

7 0

Complete Question:

Stone has been used as a building material since ancient times because of its compressive strength, which is the?

Group of answer choices

A. ability to support pressure without breaking.

B. ability to press down and become solidly fixed in the ground.

C. relative weight of a block of stone to its size.

D. force a block exerts on the blocks around it.

Answer:

A. ability to support pressure without breaking.

Explanation:

Stone has been used as a building material since ancient times because of its compressive strength, which is the ability to support pressure without breaking.

Compressive strength can be defined as the ability of a structural element or particular material to withstand an applied, which is aimed at reducing the size of the structural element.

Simply stated, it is the ability of a structural element or material to withstand an applied load without deflections, fracture or having any crack.

In this context, a stone possesses the ability to resist or withstand compression loads.

Some examples of other materials or structural elements having good compressive strength are steel, bones, concrete etc. The standard unit of measurement of the compressive strength of a material is Mega Pascal (MPa) or pound-force per square inch (psi) in the United States of America.

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The water level in a tank z1, is 20 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end

Alika [10]

Answer:

the maximum height h to which the the water steam could rise is 40.65m

Explanation:

Given;

$Z_{1}$ = 20m,

$P_{1gage}$ = 2atm ≈ 20265 N/m

Density of water ρ = 1000 kg/ $m^{2}$

Note: we take point 1 at the free surface of the water in the tank and point 2 at the top of the water trajectory. we also take reference level at the bottom of the tank.

fluid velocity at the free surface of tank is very low ( $V_{1}$ ≅ 0) and at the top of the water trajectory $V_{2}$ = 0

Step 1: Applying Bernoulli equation between poin 1 and point 2

$P_{1}$ /ρg + $V_{1} ^{2}$ /2g + $Z_{1}$ = $P_{2}$ /ρg + $V_{2} ^{2}$ /2g + $Z_{2}$

$P_{1}$ /ρg + $Z_{1}$ = $P_{atm}$ /ρg + $Z_{2}$

$Z_{2}$ = ( $P_{1}$ - $P_{atm}$ )/ρg + $Z_{1}$

Substituting values into $Z_{2}$ we have,

$Z_{2}$ = $\frac{2atm}{(1000 kg/m^{3} )(9.81 m/s^{2} )} (\frac{101325N/m^{2} }{1atm} )(\frac{1 kg.m/s^{2} }{1N} ) + 20 = 40.65 m$

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3 years ago

A paper clip is made of wire 0.75 mm in diameter. If the original material from which the wire is made is a rod 25 mm in diamete

Bas_tet [7]

Answer:

True strain = 3.7704

Explanation:

Strain is the measure an object that is stretched or deformed. This occurs when a force is applied to an object. Strain deals mostly with the change in length of the object. Strain = Δ L /L = Change in Length over the original Length:

Volume Constancy :

ΔL/L0=A0/ΔA=(D0/ ΔD)=(25mm/0.75mm)^2

ΔL/L0=44.4

Engineering strain:

Engineering strain =ΔL-L0/L0=ΔL/L0-1

Engineering strain =44.4-1=43.4

True strain, ε=In(ΔL/L0)=In(43.4)=3.7704

Note that strain has no unit, so the True strain = 3.7704

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4 years ago

An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$