Answer: Plot of
vs
would be linear with a slope of 4.
Explanation:
Given
Equation is 
Taking log both sides

It resembles with linear equation 
Here, slope of
vs
is 4.
Question:
The question is not complete. See the complete question and the answer below.
A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.
Answer:
T = 0.11029m²/sec
Radius of influence = 93.304m
expected drawdown = 3.9336m
Explanation:
See the attached file for the explanation.
Yes, the green LED emits light when we connect the banana plug wires across it.
This indicates that the threshold voltage is lower for the green LED light than the blue.
<u>Explanation</u>:
A banana plug is named for its resemblance to the shape of a banana. They are wider in the middle of the plug, and narrower at the top and bottom. The banana plug can be easily plugged or unplugged into the ports of speaker or receiver.
LED lights emit photons when it is applied with electrical charge. LED lights are more efficient and last longer than incandescent light bulbs. Green light is commonly provides the calming effect. It is generally used in hyper-pigmentation treatment.
Answer:
If a truss buckles or overturns, it is usually because of the failure of an adjacent truss or its bracing. A steel truss in a fire may buckle and overturn because of expansion or weakening from the heat. Most truss failures are the result of broken connections. Photo 1 shows a set of parallel-chord wood trusses supporting a plywood floor deck.
Explanation:
Answer:
≅ 111 KN
Explanation:
Given that;
A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8
mass = 85,000 kg
drag co-efficient (C) = 0.37
(velocity (v)= 230 m/s
density (ρ) = 1.0 kg/m³
To calculate the thrust; we need to determine the relation of the drag force; which is given as:
=
× CρAv²
where;
ρ = density of air wind.
C = drag co-efficient
A = Area of the jet
v = velocity of the jet
From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0
SO, 
We can as well say:

We can now replace
in the above equation.
Therefore,
=
× CρAv²
The A which stands as the area of the jet is given by the formula:

We can now have a new equation after substituting our A into the previous equation as:
=
× Cρ 
Substituting our data from above; we have:
=
× 
= 
= 110,990N
in N (newton) to KN (kilo-newton) will be:
= 
= 110.990 KN
≅ 111 KN
In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.