Solution :
Given :
cross sectional area = 5.6 cm square
The corresponding stress
[tex]$\sigma = \frac{F}{A}$[/tex]
[tex]$=\frac{2.5 \times 175 \ lb\times (1 \ N/ 0.2248 \ lb)}{5.6 \ cm^2 (1 \ N/100 \ cm)^2}$[/tex]
[tex]$=\frac{1946.17}{0.056 \times 10^{-2}}$[/tex]
[tex]$= 3.475 \times 10^6 \ N/m^2$[/tex]
∴ [tex]$\sigma = 3.475 \ MPa$[/tex]
And the strain is
[tex]$\epsilon = \frac{\sigma}{E}$[/tex]
[tex]$=\frac{3.475}{124 \times 10^3}$[/tex]
[tex]$= 2.80 \times 10^{-5}$[/tex]
A window‐mounted air‐conditioning unit (AC) removes energy by heat transfer from a room, and rejects energy by heat transfer to the outside air. At steady‐state, the AC cycle requires 0.434kW and has a coefficient of performance (COP) of 6.22. Determine the rate at which the energy is removed from the room air, in kW. If electricity is valued at $0.10/kw-hr, determine the cost of operating the unit for 24hrs.
Solution :
Given :
The power of the air‐conditioning (AC) unit is , W = 0.434 kW
The coefficient of performance or the COP of the air‐conditioning (AC) unit is given by = 6.22
Therefore he heat removed is given by , [tex]$Q_H = 6.22 \times 0.434$[/tex]
[tex]$Q_H = 2.7 \ kW $[/tex]
Now if the electricity is valued at 0.10 dollar per kW hour, then the operating cost of the air conditioning unit in 24 hours is given by = 0.10 x 2.7 x 24
= 6.48
Therefore the operating cost = $ 6.48 for 24 hours.
A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It has a specific gravity of solids of 2.70. Compute its moisture content and degree of saturation before it was placed in the oven.
Answer:
a) the moisture content before it was placed in the oven is 18.18%
b) degree of saturation for soil is 72.19%
Explanation:
Given the data in the question;
Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100
so we substitute
Moisture content = [(53.3 - 45.1) / 45.1 ] × 100
= (8.2/45.1) × 100
= 18.18%
Therefore the moisture content before it was placed in the oven is 18.18%
Dry Unit Weight = dry weight / volume
Dry Unit Weight = 45.1 lb / 0.45 ft³
Dry Unit Weight = 100.22 lb/ft³
we know that;
dry unit weight = (Specific gravity × unit weight of water) / (1 + e)
we also know that; unit weight of water is 62.43 lbf/ft³
so we substitute
e = (2.70×62.43 / 100.22) - 1
e = 1.68 - 1
e = 0.68
so void ratio e = 0.68
Now we determine the degree of saturation using the equation;
degree of saturation = (Moisture content × specific gravity) / void ratio
we substitute
degree of saturation = ( 18.18% × 2.7) / 0.68
= 0.49086 / 0.68
= 0.7219 ≈ 72.19%
Therefore degree of saturation for soil is 72.19%
Which option identifies the type of engineer described in the following scenario?
Sean is an engineer whose current project is a skyscraper in Richmond, VA. He relies heavily on geometry in his research of building design.
Material
Civil
Mechanical
Chemical
Answer:
civil
Explanation:
mark be branilist
Answer:
Civil
Explanation:
civil engineering – the application of planning, designing, constructing, maintaining, and operating infrastructure while protecting the public and environmental health, as well as improving existing infrastructure that may
If Nick's average stride length is 2.7 feet, how many strides will it take him to walk to school?
Answer: how far is the school?
Explanation:
Joe, a technician, is attempting to connect two hubs to add a new segment to his local network. He uses one of his CAT5 patch cables to connect them; however, he is unable to reach the new network segment from his workstation. He can only connect to it from a workstation within that segment. Which of the following is MOST likely the problem?
A. One of the hubs is defective.
B. The new hub is powered down.
C. The patch cable needs to be a CAT6 patch cable.
D. The technician used a straight-through cable.
Answer:
Option D. is correct
Explanation:
Joe uses one of his CAT5 patch cables to connect two hubs to add a new segment to his local network. As he can only connect to it from a workstation within that segment, he is not able to reach the new network segment from his workstation.
The most problem is that the technician used a straight-through cable.
Option D. is correct.
name as much parts in a car that you know
Answer:
engine suspension brake and more
Explanation:
Consider the following statement, which is intended to create an ArrayList named a to store only elements of type Thing. Assume that the Thing class has been properly defined and includes a no-parameter constructor.
ArrayList a = /* missing code */;
Which of the following can be used to replace /* missing code */ so that the statement works as intended?
A: new Thing()
B: new ArrayList()
C: new ArrayList(Thing)
D: new ArrayList()
E: new ArrayList<>(Thing)
Answer:
new ArrayList<Thing>()
Explanation:
The syntax to declare an arrayList is:
ArrayList [var-name] = new ArrayList<data-type>()
From the question;
We understand that the variable name is: a
And the data-type is: Thing
So, the ArrayList of type Thing can be defined using:
ArrayList a = new ArrayList<Thing>();
Hence:
None of the options answers the question.
