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3 years ago

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The density of a fluid is given by the empirical equation rho 70:5 exp 8:27 107 P where rho is density (lbm/ft3 ) and P is press

ure (lbf/in2 ). (a) What are the units of 70:5 and 8:27 107?

1 answer:

sergij07 [2.7K]3 years ago

3 0

Answer:

The unit of 70.5 is lbm/ft^3

The unit of 8.27×10^7 is in^2/lbf

Explanation:

The unit of 70.5 has the same unit as density which is lbm/ft^3 because exponential is found of constant values (unitless values)

The unit of 8.27×10^7 (in^2/lbf) is the inverse of the unit of pressure P (lbf/in^2) because the units have to cancel out so a unitless value can be obtained. Exponential is found of figures with no unit

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Consider the circuit below where R1 = R4 = 5 Ohms, R2 = R3 = 10 Ohms, Vs1 = 9V, and Vs2 = 6V. Use superposition to solve for the

VladimirAG [237]

Answer:

The value of v2 in each case is:

A) V2=3v for only Vs1

B) V2=2v for only Vs2

C) V2=5v for both Vs1 and Vs2

Explanation:

In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.

Also, what the problem asks is the value V2 in each case, where:

$V_2=I_2R_2=V_{ab}$

If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.

In the first case we can use an equivalent resistance between R2 and R3:

$V_{ab}'=I_1'R_{2||3}=I_1'\cdot(\frac{1}{R_2}+\frac{1}{R_3})^{-1}$

And

$V_{S1}-I_1'R_1-I_1'R_4-I_1'R_{2||3}=0 \rightarrow I_1'=0.6A$

$V_{ab}'=I_1'R_{2||3}=3V=V_{2}'$

In the second case we can use an equivalent resistance between R2 and (R1+R4):

$V_{ab}''=I_3'R_{2||1-4}=I_3'\cdot(\frac{1}{R_2}+\frac{1}{R_1+R_4})^{-1}$

And

$V_{S2}-I_3'R_3-I_3'R_{2||1-4}=0 \rightarrow I_3'=0.4A$

$V_{ab}''=I_3'R_{2||1-4}=2V$

If we consider both batteries:

$V_2=I_2R_2=V_{ab}=V_{ab}'+V_{ab}''=5V$

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3 years ago

For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho

ad-work [718]

Answer:d

Explanation:

Given

Temperature $=200^{\circ}\approc 473 K$

Also $\gamma for air=1.4$

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no. $=\frac{V}{\sqrt{\gamma RT}}$

(a) $1000 km/h\approx 277.78 m/s$

Mach no. $=\frac{277.78}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.63

(b) $500 km/h\approx 138.89 m/s$

Mach no. $=\frac{138.89}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.31

(c) $2000 km/h\approx 555.55 m/s$

Mach no. $=\frac{555.55}{\sqrt{1.4\times 287\times 473}}$

Mach no.=1.27

(d) $200 km/h\approx 55.55 m/s$

Mach no. $=\frac{55.55}{\sqrt{1.4\times 287\times 473}}$

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3 years ago

Marat540 [252]

Answer:

Explanation:

This will be possible when setting them up in summer with a certain quantity of sag, they have already know that the cables won't be able to sag any further because of the heat. During winter, when the cables contract because of the cold weather, the sag will therefore be reduced, but much tension will not be put on the cables.

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2 years ago

A cantilever timber beam with a span of L = 4.25 m supports a linearly distributed load with maximum intensity of w0 = 5.5 kN/m.

9966 [12]

Answer:

the minimum width is b= 0.1414m = 141mm

Explanation:

]given,

L= 4.25

w₀ = 5.5kN/m,

allowable bending stress = 7MPa

allowable shear stress = 875kPa

h/b = 0.67

b = ?

for a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m,

the maximum moment, M exerted by the timber is = $\frac{w₀ L²}{9√3}[/texM = w₀ L²}/{9√3 = 99.34/15.6 =6.367kNmfor a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m, the shear force, V = [tex]\frac{w₀ L}{2}[/texV = {w₀ L}/{3} = 7.79kNfor maximum bending stress of a rectangular timber, B, = [tex]\frac{6M}{bh²}$

given h/b = 0.67, i.e h=0.67b

allowable bending stress = $\frac{6M}{bh²}$ = 7000kPa

7000 = (6*6.37)/ (b *(0.67b)² ) = 38.21/0.449b³

3080b³=38.21

b³ = 38.21/3080 = 0.0124

b = 0.232m

h=0.67b = 0.67* 0.232 = 0.155m

for allowable shear stress = (3V)/(2bh)

875 = (3V)/(2bh) = (3x 7.79)/(2xbx0.67b)

875 = 23.375/1.34b²

1172.5 b²= 23.375

b² =0.0199

b= 0.1414m

h=0.67b = 0.67* 0.1414 = 0.095m

the minimum width is b= 0.1414m = 141mm

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3 years ago

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hammer [34]

Answer:

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k = 1.4 W/m*K

3 0

3 years ago