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3 years ago

6

If a soil has e=0.72, moisture content = 12% and Gs=2.72, find the weight of water in KN/m3 to be added to make the soil saturat

ed.

1 answer:

Doss [256]3 years ago

7 0

Answer:

453 KN/m^3 of water should be added make the soil saturated.

Explanation:

As we know

Gs * w = S*e

S = Gs * w/e

S = 2.72 * 0.12/0.72 = 0.453

The amount of water to be added

0.453 *1000/10 = 453 KN/m^3

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The tank of the air compressor is subjected to an internal pressure of 90 psi. If the inner diameter of the tank is 22 in., and

Shkiper50 [21]

Answer:

б₁ = 1.98 ksi

б₂ = 1.98 ksi

Explanation:

Calculating the component of stress acting at point A using circumferential stress equation, we have;

б₁ = p*r/t

= 90*11/0.25

= 3960 psi = 3.96 ksi

Calculating the component of stress acting at point A using longitudinal stress equation, we have;

б₂ = p*r/2*t

= 90*11 /2*0.25

= 1.98 ksi

Find attached of the volume element at point A.

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4 years ago

What is arduino and for what it is used

Eva8 [605]

Answer:

Arduino es una plataforma electrónica de código abierto basada en hardware y software fácil de usar. Las placas Arduino pueden leer entradas (luz en un sensor, un dedo en un botón o un mensaje de Twitter) y convertirlo en una salida: activar un motor, encender un LED, publicar algo en línea.

Explanation:

utilizado para la construcción de proyectos electrónicos. Consiste en una placa de circuito programable física y un software, o IDE (Integrated Development Environment) que se ejecuta en su computadora, donde puede escribir y cargar el código de la computadora en la placa física.

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3 years ago

Using data in Appendix A ,calculate the number of atoms in 1 tonne of iron

maksim [4K]

<h2>Answer:</h2>

$1.0783*10^{28}(atoms).$

<h2>Explanation:</h2>

<em>Since I don't have access to "Appendix A", I'll solve the problem using data from the periodic table.</em>

<em />

<h3>1. Determine the molar mass of iron.</h3>

<em>According to the periodic table, the molar mass of iron is:</em>

55.845g/mole.

<h3>2. Convert 1 tonne to grams.</h3>

$1(tonne)*1000=1000kg\\1000kg*1000=1000000g=10^6g$

<h3>3. Apply rule of 3.</h3>

$55.845g$ ----------- $1 mole$

$10^6g$ ----------- $x$

$x=\frac{10^6*1}{55.845}=17906.7061(moles)$

<h3>4. Determine the amount of atoms.</h3>

<em>Considering that there are, approximately, </em> $6.022*10^{23}$ atoms in a mole of any element, apply another rule of 3.

1 mole --------------------- $6.022*10^{23}(atoms)$

$17906.7061(moles)$ --------------------- x

$x=\frac{17906.7061*6.022*10^{23}}{1}=1.0783*10^{28}$ .

4 0

2 years ago

Refrigerant 134a at p1 = 30 lbf/in2, T1 = 40oF enters a compressor operating at steady state with a mass flow rate of 200 lb/h a

Mazyrski [523]

Answer:

a) $\mathbf{Q_c = -3730.8684 \ Btu/hr}$

b) $\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}$

Explanation:

From the properties of Super-heated Refrigerant 134a Vapor at $T_1 = 40^0 F$ , $P_1 = 30 \ lbf/in^2$ ; we obtain the following properties for specific enthalpy and specific entropy.

So; specific enthalpy $h_1 = 109.12 \ Btu/lb$

specific entropy $s_1 = 0.2315 \ Btu/lb.^0R$

Also; from the properties of saturated Refrigerant 134 a vapor (liquid - vapor). pressure table at $P_2 = 160 \ lbf/in^2$ ; we obtain the following properties:

$h_2 = 115.91 \ Btu/lb\\\\ s_2 = 0.2157 \ Btu/lb.^0R$

Given that the power input to the compressor is 2 hp;

Then converting to Btu/hr ;we known that since 1 hp = 2544.4342 Btu/hr

2 hp = 2 × 2544.4342 Btu/hr

2 hp = 5088.8684 Btu/hr

The steady state energy for a compressor can be expressed by the formula:

$0 = Q_c -W_c+m((h_1-h_e) + \dfrac{v_i^2-v_e^2}{2}+g(\bar \omega_i - \bar \omega_e)$

By neglecting kinetic and potential energy effects; we have:

$0 = Q_c -W_c+m(h_1-h_2) \\ \\ Q_c = -W_c+m(h_2-h_1)$

$Q_c = -5088.8684 \ Btu/hr +200 \ lb/hr( 115.91 -109.12) Btu/lb \\ \\$

$\mathbf{Q_c = -3730.8684 \ Btu/hr}$

b) To determine the entropy generation; we employ the formula:

$\dfrac{dS}{dt} =\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c$

In a steady state condition $\dfrac{dS}{dt} =0$

Hence;

$0=\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c$

$\sigma _c = m( s_1 -s_2) - \dfrac{Qc}{T}$

$\sigma _c = [200 \ lb/hr (0.2157 -0.2315) \ Btu/lb .^0R - \dfrac{(-3730.8684 \ Btu/hr)}{(40^0 + 459.67^0)^0R}]$

$\sigma _c = [(-3.16 ) \ Btu/hr .^0R + (7.4667 ) Btu/hr ^0R}]$

$\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}$

5 0

3 years ago

When storing used oil, it needs to be kept in ___________ containers. A) MetalB) PlasticC) UncoveredD) Leakproof

Romashka-Z-Leto [24]

Answer:

D. Leak-proof

Explanation:

Used oils are required to be stored in closed containers or tanks which will remain closed when oil is not added or removed. Alternatively, used oils can be stored in containers labeled "Used Oil" for regulated hazardous waste storing. A leak-proof container is one that has no drips and dribbles when the container lid is closed. These containers should be kept in good conditions away from rusting, leaks or deteriorations. It is important to mention that used oil should not be stored in anything apart from tanks and leak-proof storage containers.

8 0

3 years ago