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lilavasa [31]
2 years ago
13

A manometer measures a pressure difference as 40 inches of water. Take the density of water to be 62.4 lbm/ft3.What is this pres

sure difference in pound-force per square inch, psi?

Engineering
1 answer:
Artemon [7]2 years ago
5 0

Answer:

1.444psi

Explanation:

You might be interested in
1. Add:<br>(i) 5xy, -2xy, -11xy, 8xy<br>(iv) 3a - 2b + c, 5a + 8b -70​
Cerrena [4.2K]

Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

Hope this helps you

8 0
1 year ago
Your task is to fill in the missing parts of the C code to get a program equivalent to the generated assembly code. Recall that
Rudik [331]

Answer:

See Explaination

Explanation:

//Function

long loop (long x, long n)

{

//Declare a variable named result and initialize it to zero

long result = 0;

//Declare a variable named mask

long mask;

//For loop

for(mask = 1; mask != 0; mask = mask << (n & 0xFF))

{

//Calculate

result | = (x&mask);

}

//Return result

return result;

}

6 0
2 years ago
Liquid oxygen is stored in a thin-walled, spherical container 0.75 m in diameter, which is enclosed within a second thin-walled,
nadezda [96]

Answer:

Explanation:

the solution is well stated

3 0
2 years ago
A four-variable logic function that is equal to 1 if any three or all four of its variables are equal to 1 is called a majority
gulaghasi [49]

Answer:

1. Assumption

2.Truth Table

3.K-Map

4.Mapping and Final Expression

5 0
2 years ago
1.00-L insulated bottle is full of tea at 90.08°C. You pour out one cup of tea and immediately screw the stopper back on the bot
liberstina [14]

Answer:

T_{f} = 90.07998 ° C

Explanation:

This is a calorimetry process where the heat given by the Te is absorbed by the air at room temperature (T₀ = 25ºC) with a specific heat of 1,009 J / kg ºC, we assume that the amount of Tea in the cup is V₀ = 100 ml. The bottle being thermally insulated does not intervene in the process

                 Qc = -Qb

                M c_{e_Te} (T₁ -T_{f}) = m c_{e_air} (T_{f}-T₀)

Where M is the mass of Tea that remains after taking out the cup, the density of Te is the density of water plus the solids dissolved in them, the approximate values are from 1020 to 1200 kg / m³, for this calculation we use 1100 kg / m³

   ρ = m / V  

   V = 1000 -100 = 900 ml  

   V = 0.900 l (1 m3 / 1000 l) = 0.900 10⁻³ m³  

   V_air = 0.100 l = 0.1 10⁻³ m³  

Tea Mass  

     M = ρ V_te  

     M = 1100 0.9 10⁻³  

     M = 0.990 kg  

Air mass  

     m = ρ _air V_air  

     m = 1.225 0.1 10⁻³  

     m = 0.1225 10⁻³ kg  

(m c_{e_air} + M c_{e_Te}) T_{f}. = M c_{e_Te} T1 - m c_{e_air} T₀  

T_{f} = (M c_{e_Te} T₁ - m c_{e_air} T₀) / (m c_{e_air} + M c_{e_Te})  

Let's calculate  

T_{f} = (0.990 1100 90.08– 0.1225 10⁻³ 1.225 25) / (0.1225 10⁻³ 1.225 + 0.990 1100)  

T_{f} = (98097.12 -3.75 10⁻³) / (0.15 10⁻³ +1089)  

T_{f} = 98097.11 / 1089.0002  

T_{f} = 90.07998 ° C  

This temperature decrease is very small and cannot be measured

3 0
2 years ago
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