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.
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%
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.
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
name as much parts in a car that you know
Answer:
engine suspension brake and more
Explanation:
