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coldgirl [10]
3 years ago
13

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

a building are 730 and 755 mmHg, respectively, determine the height of the building. Assume an average air density of 1.18 kg/m^3.
Engineering
1 answer:
bixtya [17]3 years ago
7 0

Answer:

h = 287.1 m

Explanation:

the density of mercury \rho =13570 kg/m3

the atmospheric pressure at the top of the building is

p_t = \rho gh  = 13570*908*0.73 = 97.08 kPa

the atmospheric pressure at bottom

p_b = \rho gh  = 13570*908*0.75 = 100.4 kPa

\frac{w_{air}}{A} =p_b -p_t

we have also

(\rho gh)_{air} = p_b - p_t

1.18*9.81*h = (100.4 -97.08)*10^3

h = 287.1 m

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What is the area enclosed by the cycle area of the Carnot cycle illustrating on a P-V diagram?
Inga [223]

Answer:

The work of the cycle.

Explanation:

The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.

The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.

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12.50 An air conditioner operating at steady state takes in moist air at 28°C, 1 bar, and 70% relative humidity. The moist air f
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3 years ago
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6 0
3 years ago
Problem 34.3 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E
Natasha2012 [34]

The image is missing, so i have attached it.

Answer:

A) P = 65.11 KN

B) Q = 30 KN

Explanation:

We are given;

The end reaction of the beam; F = 100KN

Coefficient of static friction between two steel surfaces;μ_ss = 0.3

Coefficient of static friction between steel and concrete;μ_sc = 0.6

So, F1 = μ_ss•F =0.3 x 100 = 30 KN

F2 = μ_ss•N_EF = 0.3N_EF

From the screen shot, we see that the angle is 12°

Sum of forces in the Y-direction gives;

F2•sin12 - N_EF•cos12 + 100 = 0

Rearranging gives;

N_EF•cos12 - F2•sin12 = 100

Let's put 0.3N_EF for F2 to give;

N_EF•cos12 - 0.3N_EF•sin12 = 100

Thus;

N_EF(0.9158) - 0.1247 = 100

N_EF(0.9781) = 100 + 0.1247

N_EF = 100.1247/0.9158

N_EF = 109.33 KN

Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN

Wedge will move if;

P = (F1 + F2cos12 + N_EFsin12)

Thus;

P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)

P ≥ 65.11 KN

B) For static equilibrium, Q = F1

Thus, Q = 30 KN

3 0
3 years ago
The modulus of elasticity for a ceramic material having 6.0 vol% porosity is 303 GPa. (a) Calculate the modulus of elasticity (i
Phantasy [73]

Answer:

modulus of elasticity for the nonporous material is 340.74 GPa

Explanation:

given data

porosity = 303 GPa

modulus of elasticity = 6.0

solution

we get here  modulus of elasticity for the nonporous material Eo that is

E = Eo (1 - 1.9P + 0.9P²)    ...............1

put here value and we get Eo

303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )  

solve it we get

Eo = 340.74 GPa

8 0
3 years ago
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