The option that is not associated with the given term called urban sprawl is; Option A: Blocking high views
What is Urban Sprawl?
Urban sprawl is defined as the rapid expansion of the geographic boundaries of towns and cities which is often accompanied by low-density residential housing and increased reliance on the private automobilefor movement.
Looking at the given options, "blocking high views" is the option that is not typically a problem associated with urban sprawl because urbanization usually takes place on relatively flat levels.
The missing options are;
a. blocking high views
b. destroying animal habitats
c. overrunning farmland
d. reducing green space
Read more about urban sprawl at; brainly.com/question/504389
Complete Question
The complete question is shown on the first uploaded image
Answer:
a) 
b) 
Explanation:
The explanation is shown on the second and third uploaded image
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.
Answer:
A.) Find the answer in the explanation
B.) Ua = 7.33 m/s , Vb = 7.73 m/s
C.) Impulse = 17.6 Ns
D.) 49%
Explanation:
Let Ua = initial velocity of the rod A
Ub = initial velocity of the rod B
Va = final velocity of the rod A
Vb = final velocity of the rod B
Ma = mass of rod A
Mb = mass of rod B
Given that
Ma = 2kg
Mb = 1kg
Ub = 3 m/s
Va = 0
e = restitution coefficient = 0.65
The general expression for the velocities of the two rods after impact will be achieved by considering the conservation of linear momentum.
Please find the attached files for the solution
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure =
= 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ