The assignment is to modify the class implementation without changing the functionality. Remove the feet and inches data members
and replace them with one data member named total_inches. Re-implement all the class methods so they have he same functionality as originally. This means that any program written using the public members of the new class must produce exactly the same results as the original implementation.
1 answer:
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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
Answer:
Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.
Explanation:
Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow. You can also see that represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:
- Wall Thickness. More thickness, more thermal resistance.
- Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
- Cross-section Area. More cross-section area, less thermal resistance.
A expression to define the thermal resistance for the wall is as follows: , where l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.
Answer:
A) i) 984.32 sec
ii) 272.497° C
B) It has an advantage
C) attached below
Explanation:
Given data :
P = 2700 Kg/m^3
c = 950 J/kg*k
k = 240 W/m*K
Temp at which gas enters the storage unit = 300° C
Ti ( initial temp of sphere ) = 25°C
convection heat transfer coefficient ( h ) = 75 W/m^2*k
<u>A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>
First step determine the Biot Number
characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125
Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3
Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method
attached below is a detailed solution of the given problem
<u>B) The physical properties are copper</u>
Pcu = 8900kg/m^3)
Cp.cu = 380 J/kg.k
It has an advantage over Aluminum
C<u>) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>
Given that:
P = 2200 Kg/m^3
c = 840 J/kg*k
k = 1.4 W/m*K
