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Vlada [557]
4 years ago
8

You are given a body with no body forces and told that the stress state is given as: ⎡ ⎣ 3αx 5βx2 + αy γz3 5βx2 + αy βx2 0 γz3 0

5 ⎤ ⎦ psi, where (α, β, γ) are constants with the following values: α = 1 psi/in, β = 1 psi/in2, and γ = 1 psi/in3. Does this represent an equilibrium state of stress? Assume the body occupies the domain Ω = [0, 1] × [0, 1] × [0, 1] (in inches).

Engineering
1 answer:
Lady bird [3.3K]4 years ago
5 0

Answer:

This doesn't represent an equilibrium state of stress

Explanation:

∝ = 1 , β = 1 ,  y = 1

x = 0 , y = 0 , z = 0 ( body forces given as 0 )

Attached is the detailed solution is and also the conditions for equilibrium

for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution

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A flywheel performs each of these functions except: A. Contains a gear used for engine starting B. Smoothes engine operation C.
torisob [31]

Answer:

C. Provides lubrication to parts

Explanation:

Flywheel :

 Flywheel is a device which stored the mechanical energy.This energy can be use when more energy required during any operation.Due to high moment of inertia of the flywheel it resist the change in the speed .The flywheel is attached to the crank shaft at the rear side of the engine.

The flywheel perform following function:

1. It connects the crankshaft and the transmission system.

2.It makes the engine operation smooth.

3.It contains gear and other parts of the engine.

But it can not provide lubrication.

C. Provides lubrication to parts

4 0
3 years ago
4. Two technicians are discussing the evaporative emission monitor. Technician A says that serious monitor faults cause a blinki
snow_lady [41]

Answer:

The correct option is;

Neither Technician A nor B

Explanation:

The evaporative emission monitor or Evaporaive Emission Control System EVAP System monitors enables the Power Control Module of the car to check fuel system leak integrity and the vapor consumption efficiency during engine combustion

It is a requirement of EPA on cars to check the emission of smug forming evaporates from cars

Serious monitor faults can cause the turning on of the check engine lights and the vehicle will not pass OBD II test, but it will not lead to engine shutdown

It runs when the engine is 15 to 85% full and the TP sensor is between 9% and 35%.

Therefore, the correct option is that neither Technician A nor B are correct.

3 0
3 years ago
A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th
Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ    

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588  = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

 cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 =  2552.64 kJ    

b)  \Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)

Where:

x_A and x_B are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, x_A = 248 /(248 + 42.8) = 0.83

x_B = 42.8/(248 + 42.8) = 0.1472

∴ \Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472) =  1066.0279 J/K

5 0
3 years ago
The input and output signals of a system is related by the following equation: fraction numerator d squared y over denominator d
Colt1911 [192]

Answer:

Explanation:

The given equation is :

\frac{d^{2}y }{dx^{2} } + sin(3y) \frac{dy}{dt} + y = t\frac{df}{dt} + f

5 0
3 years ago
100 kg of R-134a at 200 kPa are contained in a piston–cylinder device whose volume is 12.322 m3. The piston is now moved until t
LekaFEV [45]

Answer:

T=151 K, U=-1.848*10^6J

Explanation:

The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:

 pV=nRT

For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.  

Using the given mass m, molar mass M, we can get the following equation:  

 pV=mRT/M

To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:

T=pVM/(Rm); so initial T=302.61K and final T=151.289K

 

Now we can calculate change of U:

U=3/2 mRT/M using T- difference in temperatures

 U=-1.848*10^6 J

Note, that the energy was taken away from the system.  

5 0
4 years ago
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