Answer:
1. Equatorial Evergreen or Rainforest
2. Tropical forest
3. Mediterranean forest
4. Temperate broad-leaved forest
5. Warm temperate forest
Explanation:
Answer:
hello your question lacks the required diagram attached below is the diagram
answer : 58.47 N
Explanation:
The magnitude of the force acting on the Pin D
Fd = 
= 
= 58.465 N
Dx = 16.80 N
Dy = 56 N
hello attached below is the detailed solution
Answer:
I think D is correct
Explanation:
C is decreasing function, probably worst
A is arctan -> in radian, the rate of increasing is very slow-> second worst
B(14) = ln(9*14) = 4.8
D(14) = sqrt(8+14^2)=14.2
Answer:
2800 [MPa]
Explanation:
In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

Where
is the magnitude of he maximum stress at the tip of the crack,
is the magnitude of the tensile stress,
is
the length of the internal crack, and
is the radius of curvature of the crack.
We have:
![r_{c}=1.9*10^{-4} [mm]](https://tex.z-dn.net/?f=r_%7Bc%7D%3D1.9%2A10%5E%7B-4%7D%20%5Bmm%5D)
![l_{c}=3.8*10^{-2} [mm]](https://tex.z-dn.net/?f=l_%7Bc%7D%3D3.8%2A10%5E%7B-2%7D%20%5Bmm%5D)
![\sigma_{c}=140 [MPa]](https://tex.z-dn.net/?f=%5Csigma_%7Bc%7D%3D140%20%5BMPa%5D)
We replace:
![\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}](https://tex.z-dn.net/?f=%5Csigma_%7Bm%7D%20%3D%202%2A%28140%20%5BMPa%5D%29%2A%28%5Cfrac%7B%5Cfrac%7B3.8%2A10%5E%7B-2%7D%20%5Bmm%5D%7D%7B2%7D%7D%7B1.9%2A10%5E%7B-4%7D%20%5Bmm%5D%7D%29%5E%7B0.5%7D)
We get:
![\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}=2800 [MPa]](https://tex.z-dn.net/?f=%5Csigma_%7Bm%7D%20%3D%202%2A%28140%20%5BMPa%5D%29%2A%28%5Cfrac%7B%5Cfrac%7B3.8%2A10%5E%7B-2%7D%20%5Bmm%5D%7D%7B2%7D%7D%7B1.9%2A10%5E%7B-4%7D%20%5Bmm%5D%7D%29%5E%7B0.5%7D%3D2800%20%5BMPa%5D)
Answer:
Space mean speed = 44 mi/h
Explanation:
Using Greenshield's linear model
q = Uf ( D -
/Dj )
qcap = capacity flow that gives Dcap
Dcap = Dj/2
qcap = Uf. Dj/4
Where
U = space mean speed
Uf = free flow speed
D = density
Dj = jam density
now,
Dj = 4 × 3300/55
= 240v/h
q = Dj ( U -
/Uf)
2100 = 240 ( U -
/55)
Solve for U
U = 44m/h