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love history [14]
4 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]4 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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Which of the following is an example of a new and emerging job in the green economy?(Brainliest for who answers)
grigory [225]

Answer: Fergie is a consultant specializing in deriving commercial power from waves and tides.

Explanation:

The green economy the economy that has the goal of minimizing environmental risks while aiming for the sustainable development of the economy without having a negative effect on the environment.

According to the Bureau of Labor Statistics, some categories of Green Jobs are Water conservation, environmental remediation, wind energy, solar power, energy auditors, recycling etc.

Out of the options given in the question, an example of a new and emerging job in the green economy is option D "Fergie is a consultant specializing in deriving commercial power from waves and tides".

5 0
3 years ago
Read 2 more answers
The equation for the velocity V in a pipe with diameter d and length L, under laminar condition is given by the equation V=Δpdsq
Citrus2011 [14]

Answer:

Given that

V=\dfrac{\Delta Pd^2}{32\mu L}

LHS of above given equation have dimension [M^oL^{1}T^{-1}].

Now find the dimension of RHS

Dimension of P = [ML^{-1}T^{-2}].

Dimension of d=  [M^{0}L^{1}T^{0}].

Dimension of μ =  [ML^{-1}T^{-1}].

Dimension of L=  [M^{0}L^{1}T^{0}].

So

\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}

\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]

It means that both sides have same dimensions.

5 0
4 years ago
A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.
Alexus [3.1K]

Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

Explanation:

First, let us define the parameters in the question:

void ratio e  = \frac{V_v}{V_s} =  \frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)

Specific gravity G_{s} = \frac{P_s}{P_w} =  \frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)

% Saturation S = \frac{V_w}{Vv} × \frac{100}{1} =  \frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right} × \frac{100}{1}--------(3)

water content w =  \frac{M_w}{M_s} = \frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w =  \frac{M_w}{Ms}

Recall; mass = Density × volume

w = \frac{V_wP_w}{V_sP_s} ------(5)

From eqn. (2)  G_{s} = \frac{P_s}{P_w}

∴ \frac{1}{G_s} = \frac{P_w}{P_s} ------(6)

putting eqn. (6) into (5)

w = \frac{V_w}{V_sG_s} -----(7)

Again, from eqn (1)

V_s = \frac{V_v}{e}

substituting into eqn. (7)

w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v}  = S

∴ w = \frac{Se}{G_s} -----(8)

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, G_{s} = 2.7, e= 0.9

∴w= \frac{0.9*1}{2.7} = 0.333

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

w= \frac{0.01*0.9 }{2.7} = 0.0033

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil (P_w=\frac{1_g}{cm^{3} } )

Given, V_{s} = 100 cm^{3 }, S = 50% = 0.5

%S = \frac{V_w}{V_v} × \frac{100}{1} = \frac{V_w}{eV_s} × \frac{100}{1}

0.50 = \frac{V_w}{0.9* 100}  = 45cm^{3}

mass of water = P_wV_w= 1 * 45 = 45_{g}

7 0
4 years ago
1. The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle.
sammy [17]

Answer:

A. Yes

B. Yes

Explanation:

We want to evaluate the validity of the given assertions.

1. The first statement is true

The sine rule stipulates that the ratio of a side and the sine of the angle facing the side is a constant for all sides of the triangle.

Hence, to use it, it’s either we have two sides and an angle and we are tasked with calculating the value of the non given side

Or

We have two angles and a side and we want to calculate the value of the side provided we have the angle facing this side in question.

For notation purposes;

We can express the it for a triangle having three sides a, b, c and angles A,B, C with each lower case letter being the side that faces its corresponding big letter angles

a/Sin A = b/Sin B = c/Sin C

2. The cosine rule looks like the Pythagoras’s theorem in notation but has a subtraction extension that multiplies two times the product of the other two sides and the cosine of the angle facing the side we want to calculate

So let’s say we want to calculate the side a in a triangle of sides a, b , c and we have the angle facing the side A

That would be;

a^2 = b^2 + c^2 -2bcCosA

So yes, the cosine rule can be used for the scenario above

3 0
3 years ago
Two very long concentric cylinders of diameters D1 = 0.42 m and D2 = 0.5 m are maintained at uniform temperatures of T1 = 950 K
MrRa [10]

Answer:

Q=33.34 KW/m

Explanation:

Given that

D₁=0.42 m

A₁= π D₁ L

For unit length

A₁= π D₁ = 0.42 π m²

D₂=0.5 m

A₂= 0.5 π m²

ε₁= 1 ,ε₂= 0.55

T₁=950 K  ,T₂ = 500 K

Q=\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{A_1\epsilon _1}+\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+\dfrac{1}{A_1F_{12}}}

F₁₁+F₁₂= 1

F₁₁= 0

So, F₁₂= 1

Q=A_1\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{\epsilon _1}+A_1\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+1}

Q=0.42\pi \dfrac{5.67\times 10^{-8}(950^4-500^4)}{\dfrac{1-1}{1}+0.42\pi\times \dfrac{1-0.55}{0.5\pi\times 0.55}+1}

Q=33.34 KW/m

3 0
4 years ago
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