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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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An organization is struggling to differentiate threats from normal traffic and access to systems. A security engineer has been a
AnnZ [28]

Answer:

Web application Firewall (WAF)

Explanation:

The Web application Firewall (WAF) will be recommended. This will enormously help any organisation that is having a struggle of differentiating threats from normal traffic and access to systems.

WAF is also going to It aggregate data and provide metrics that will assist in identifying malicious actors. Another vital function of WAF will be to block unwanted web traffic from accessing your site. It will protect against hacks, brute force attacks, DDoS attacks, cross-site scripting, SQL injection, and zero-day exploits.

8 0
4 years ago
Two piezometers have been placed along the direction of flow in a confined aquifer that is 30.0 m thick. The piezometers are 280
Dafna11 [192]

Answer:

time = 224 days

Explanation:

given data

thick = 30 m

piezometers =  280 m

head between  two = 1.4 m

aquifer hydraulic conductivity = 50 m

porosity =  20%

solution

we get here average pole velocity that is get by using Darcy law that is

Va = \frac{k}{\eta } \times \frac{\Delta h}{L}   ................1

here Va is average pole velocity and k is hydraulic conductivity and \eta is porosity  

here v is = k \times  \frac{dh}{dl}   ...........2

v = 50 × \frac{1.4}{280}

v = 0.25 m/day

and here average linear velocity Va will be

Va = \frac{v}{\eta }  

Va = \frac{0.25}{0.2}  

Va = 1.25 m/day  

travel time for water will be

time = \frac{280}{1.25}  

time = 224 days

8 0
3 years ago
Which of the following should NOT be included in an emergency kit?
Jet001 [13]

Answer:

Pistol

Explanation: would not be used to help in an emergency kit

3 0
3 years ago
Read 2 more answers
As shown, a load of mass 10 kg is situated on a piston of diameter D1 = 140 mm. The piston rides on a reservoir of oil of depth
telo118 [61]

Answer:

165 mm

Explanation:

The mass on the piston will apply a pressure on the oil. This is:

p = f / A

The force is the weight of the mass

f = m * a

Where a in the acceleration of gravity

A is the area of the piston

A = π/4 * D1^2

Then:

p = m * a / (π/4 * D1^2)

The height the oil will raise is the heignt of a colum that would create that same pressure at its base:

p = f / A

The weight of the column is:

f = m * a

The mass of the column is its volume multiplied by its specific gravity

m  = V * S

The volume is the base are by the height

V = A * h

Then:

p = A * h * S * a / A

We cancel the areas:

p = h * S * a

Now we equate the pressures form the piston and the pil column:

m * a / (π/4 * D1^2) = h * S * a

We simplify the acceleration of gravity

m / (π/4 * D1^2) = h * S

Rearranging:

h = m / (π/4 * D1^2 * S)

Now, h is the heigth above the interface between the piston and the oil, this is at h1 = 42 mm. The total height is

h2 = h + h1

h2 = h1 + m / (π/4 * D1^2 * S)

h2 = 0.042 + 10 / (π/4 * 0.14^2 * 0.8) = 0.165 m = 165 mm

7 0
3 years ago
Why does my man bun not have its own erodynamics
Aloiza [94]

Answer:

umm okay for starters I have no clue lol.

7 0
3 years ago
Read 2 more answers
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