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Darya [45]
3 years ago
11

Atmospheric pressure is measured to be 14.769 psia. a. What would be the equivalent reading of a water barometer (inches of H20)

? b. What would be the equivalent reading of a Mercury barometer (mm of Hg)?
Engineering
1 answer:
Fofino [41]3 years ago
6 0

Answer:

(a) water height =408.66 in.

(b) mercury height=30.04 in.

Explanation:

Given: P=14.769 psi     ( 1 psi= 6894.76 \frac{N}{m^2} )

we know that   P=\rho\times g\timesh

where \rho =Density,g=9.81\frac{m}{s^2}

     h=height.

Given that P=14.769 psi ⇒P= 101828.6 7\dfrac{N}{m^2}

(a) P=\rho_{w}\times g\times h_{w}  

     \rho_{w}=1000\frac{Kg}{m^3}

⇒101828.67=1000\times 9.81\times h_{w}

h_{w}=10.38 m

So water barometer will read 408.66 in.            (1 m=39.37 in)

(b)  P=\rho_{hg}\times g\times h_{hg}

     \rho_{hg}=13600

So 101828.67=13600\times 9.81\times h_{hg}

h_{hg}=0.763 m

So mercury barometer will read 30.04 in.

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kakasveta [241]

Answer:

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Explanation:

using given data to obtain values from table F7.1

Enthalpy of water at temperature of 100 F = 68.04Btu/Ibm

Enthalpy of water at temperature of 50 F = 18.05 Btu/Ibm

from table F.3

specific constant of glycerin C_{p} = 0.58 Btu/Ibm-R

<u>The rate of internal heat transfer ( change in enthalpy ) </u>

h4 - h3 = Cp ( T4 - T3 ) --------------- ( 1 )

where ; T4 = 50 F

             T3 = 10 F

             Cp = 0.58 Btu/Ibm-R

substitute given values into equation 1

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<u>Determine mass flow rate of glycol</u>

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3 years ago
Problem 3: Soil Classification using the AASHTO and USCS Systems
nataly862011 [7]

<u>Solution:</u>

Given\\                   \(\quad W=3000 Ib , \quad m=\frac{W}{g}=\frac{3000}{322} \ slug =93.1677 slug\)\\K_{e q}=2160 lbs / wp =2100 \frac{ lbs }{10} \frac{ x 12}{1 ft }=(2160 \times 12) lb / ft$$

a) The natural frequency

\begin{aligned}&\left(\omega_{n}\right)=\sqrt{\frac{K_{e q}}{m}}\\&=\sqrt{\frac{2160 \times 12}{93.1677}}\\&\omega_{n}=16.68 \text { rad } | s\\&\omega_{n}=\frac{2 \pi}{T}\\&16.68=\frac{2 \pi}{T}\\&T=0.3767 s\end{aligned}

b)

Given, \(t=10 s , \quad y(t)=6 in = A\)\\\(y(t)=A \cos \left(\omega_{n} t+\phi\right) \rightarrow 0\)\\\(6=6 \cos (16.68 \times 10+\phi)\)\\\(1=\cos (166.8+\phi)\)\\\(166.8+\phi=0\)\\\phi=-166.8\)\\At \(t=0, \quad y(0)=6 \cos (16.68 \times 0-166.8)\) {y(0)}=-5.74 in

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3 years ago
Which branch of engineering most closely relates to mechanical engineering?
Rus_ich [418]

Answer:

C

Explanation:

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Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f
natita [175]

Answer:

Complete question is:

write the following decorators and apply them to a single function (applying multiple decorators to a single function):

1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <strong> and </strong> to the argument of the decorator. The return value of the wrapper should look like: return “<strong>” + func() + “</strong>”

2. The decorator will return the wrapper per usual.

3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <em> and </em> to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “<em>” + func() + “</em>.

4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.

5. Apply both decorators (by @ operator to greetings()).

6. Invoke the greetings() function and capture the result.

Code :

def strong_decorator(func):

def func_wrapper(name):

return "<strong>{0}</strong>".format(func(name))

return func_wrapper

def em_decorator(func):

def func_wrapper(name):

return "<em>{0}</em>".format(func(name))

return func_wrapper

@strong_decorator

@em_decorator

def Greetings(name):

return "{0}".format(name)

print(Greetings("Hello"))

Explanation:

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