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love history [14]
3 years ago
11

One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacu

um. The two sides are separated by a piston that is initially held in place by the pins. The pins are removed and the gas suddenly expands until it hits the stops. What happens to the internal energy of the gas?
a. internal energy goes up
b. internal energy goes down
c. internal energy stays the same
d. we need to know the volumes to make the calculation
Engineering
1 answer:
laiz [17]3 years ago
7 0

Answer:

Option C = internal energy stays the same.

Explanation:

The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.

So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.

Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.

The amount of heat,q = Work,w.

In the concept of free expansion the only thing that changes is the volume.

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Answer:

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4 0
2 years ago
Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a
Elena L [17]

Answer:

\mathbf{\tau_c =5.675 \ MPa}

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [\bar 101]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]

where;

[d_1\ e_1 \ f_1] = directional indices for tensile stress

[d_2 \ e_2 \ f_2] = slip direction

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_2 = -1 , e_2 = 0 , f_2 = 1

cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]

cos \ \lambda = \dfrac{1}{\sqrt{2}}

Also, to find the angle \phi between the stress [001] & normal slip plane [111]

Then;

cos \  \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_3 = 1 , e_3 = 1 , f_3 = 1

cos \  \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]

cos \phi= \dfrac{1} {\sqrt{3} }

However, the critical resolved SS(shear stress) \mathbf{\tau_c} can be computed using the formula:

\tau_c = (\sigma )(cos  \phi )(cos \lambda)

where;

applied tensile stress \sigma = 13.9 MPa

∴

\tau_c =13.9\times (  \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})

\mathbf{\tau_c =5.675 \ MPa}

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3 years ago
In software engineering how do you apply design for change?
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Answer:

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7 0
3 years ago
Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per
kirill [66]

Solution :

Given :

$V_1 = 7 \ m/s$

Operation time, $T_1$ = 3000 hours per year

$V_2 = 10 \ m/s$

Operation time, $T_2$ = 2000 hours per year

The density, ρ = $1.25 \ kg/m^3$

The wind blows steadily. So, the K.E. = $(0.5 \dot{m} V^2)$

                                                             $= \dot{m} \times 0.5 V^2$

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = $\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$

Regarding that $\dot m \propto V$. Then,

Power $ \propto V^3$ → Power = constant x $V^3$

Since, $\rho_a$ is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, $P_1= \text{const.} \times V_1^3$

            $P_1 = \text{const.} \times 343 \ W$

For the second site,

Power, $P_2 = \text{const.} \times V_2^3 \ W$

           $P_2 = \text{const.} \times 1000 \ W$

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A double-threaded Acme stub screw of 2-in. major diameter is used in a jack having a plain thrust collar of 2.5-in. mean diamete
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This is the answer for the question

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