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3 years ago

The "Big Dig" was the nickname of the civil engineering project that redesigned the highway Infrastructure for the city of

Engineering

1 answer:

zheka24 [161]3 years ago

4 0

Geotechnical since it’s geologicaly based

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g Asbestos is a fibrous silicate mineral with remarkably high tensile strength. But is no longer used because airborne asbestos

otez555 [7]

Answer:

Aluminum cross sectional area = 1.99 * 10^-5 mm^2

Steel cross sectional area = 9.95* 10^-6 mm^2

Explanation:

Given data:

Tensile strength of Grunerite = 3.5 * 10^2 kg/mm^2 = 3.5 * 10^-4 kg/um^2

Tensile strength of Aluminum = 2.5 × 10^4 lb/in2 = 2.5 × 10^4 * 703.07 kg/m^2

Tensile strength of Steel number 5137 = 5.0 × 10^4 lb/in2 = 5.0 × 10^4*703.07 kg/m^2

Calculating the cross sectional area of the wires of aluminum and steel No5137

first we will determine the cross sectional area of the aluminum wire ( A ) by equating tensile strength of aluminum with the tensile strength

of Grunerite

(2.5 × 10^4 * 703.07) * A = 3.5 * 10^-4 kg

Hence ; A = 1.99 * 10^-5 mm^2

Next we calculate the cross sectional area of steel

5.0 × 10^4*703.07 kg/m^2 * A" = 3.5 * 10^-4 kg

Hence ; A" = 9.95* 10^-6 mm^2

8 0

2 years ago

A life cycle assessment (LCA) determines the environmental impact at all stages of a product's life cycle, including production,

lisabon 2012 [21]

Confusing very very confusing

4 0

3 years ago

The button on the _ valve should be held when pressure bleeding the brakes

Kamila [148]

Answer:

Meter would be your answer, hope this helped

4 0

3 years ago

Technician A says that the starter solenoid switches the high current on and off. Technician B says that the solenoid on the sta

jek_recluse [69]

Technician A is incorrect so that means Technician B is correct.

3 0

2 years ago

A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the

Sergeu [11.5K]

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

By definition the entropy change would be defined as

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})$

Using the Boyle equation we have

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})$

Where,

$C_p$ = Specific heat at constant pressure

$T_1$ = Initial temperature of gas

$T_2$ = Final temprature of gas

R = Universal gas constant

$v_1$ = Initial specific Volume of gas

$v_2$ = Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

$T_1 = T_2, v_2=v_1$

The equation would turn out as

$\Delta S = C_p ln1-ln1$

Therefore the entropy change of the ideal gas is 0

Into the surroundings we have that

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

$\Delta S = \frac{230kJ}{(30+273)K}$

$\Delta S = 0.76 kJ/K$

Therefore the increase of entropy into the surroundings is 0.76kJ/K

8 0

2 years ago