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3 years ago

Please help ASAP!!

Engineering

2 answers:

masha68 [24]3 years ago

8 0

Answer:

Explanation:

Tems11 [23]3 years ago

4 0

Answer:

a. Falls

Explanation:

plato

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What engine does the mercedes 500e have

Strike441 [17]

Answer:

5.0 m119

Explanation:

It has the 5.o m119 v8 engine which produced 322 bhp.

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2 years ago

Thermodynamics deals with the macroscopic properties of materials. Scientists can make quantitative predictions about these macr

Arlecino [84]

Answer:

Explanation:

From the information given:

Using the equipartition theorem, the average energy of a molecule dor each degree of freedom is:

$U = \dfrac{1}{2}k_BT$

$U = \dfrac{1}{2}nRT$

For s degree of freedom

$U = \dfrac{1}{2}snRT$

However, the molar specific heat $C_v = \dfrac{1}{n} \dfrac{dU}{dT}$

Therefore, in terms of R and s;

$C_v = \dfrac{1}{n} \dfrac{d}{dT} \begin{pmatrix} \dfrac{1}{2} snRT \end {pmatrix}$

$C_v = \dfrac{Rs}{2}$

Given that:

Cv=70.6Jmol⋅K and R=8.314Jmol⋅K

Then; using the formula $C_v = \dfrac{Rs}{2}$

$70.6 \ J/mol.K = \dfrac{(8.314 \ J/mol.K)\times s}{2}$

$70.6 \ J/mol.K \times 2= (8.314 \ J/mol.K)\times s$

$s= \dfrac{70.6 \ J/mol.K \times 2}{ (8.314 \ J/mol.K) }$

s = 16.983

s $\simeq$ 17

7 0

3 years ago

Two kg of N2 at 450 K, 7 bar is contained in a rigid tank connected by a valve to another rigid tank holding 1 kg of O2 at 300 K

11111nata11111 [884]

Answer:WHo knows.Explanation:

7 0

2 years ago

If link AB of the four-bar linkage has a constant counterclockwise angular velocity of 58 rad/s during an interval which include

katrin2010 [14]

Answer:Vb=-6i-(-0.1ωab+8)j m/s

Explanation:

Va=V0+Va0

Va=V0+(ra0 x ωao)

ω=Angular velocity of link A0

Using r0a=0.1m;

Va=V0+(0.1i x ω0a K)

Va=0

ixk=j

Va=0+0.1ω0aj

Calculating te velocity of using te equation below

Vb=Va+Vba

Vb=Va+ωab x rba

ωab=40rad/s

rab=-0.21i+0.15j

Va=0.1ω0aj

Vb=Va+ωabxrba

Vb=0.1ω0aj+40k x -(0.21i+0.15j)

Vb=0.1ω0aj-8j-6i

Vb=-6i-(-0.1ωab+8)j m/s

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3 years ago

One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp

pentagon [3]

Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

$m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}$

1x0.723x $(T_{eq} -350)$ =3x0.780x $(450-T_{eq} )$ ⇒ $T_{eq}$ = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

$V_{a,i} = \frac{mRT_{a,i} }{P_{a,i} }$ = $\frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar}$ = 0.201 $m^{3}$

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

$P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }$

$P_{eq}$ = 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

(0.201+1.275)

= 246.67 KPa = 2.47 bar

6 0

2 years ago