Answer:
R = ![\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%26cos30%26-sin30%5C%5C0%26sin30%26cos30%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%2060%26-sin60%260%5C%5Csin60%26cos60%2660%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
Explanation:
The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:
Let
and
be the the angles 60⁰ and 30⁰ respectively
that is
= 60⁰ and
= 30⁰
The matrix is given by the following expression:
![\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%26cos30%26-sin30%5C%5C0%26sin30%26cos30%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%2060%26-sin60%260%5C%5Csin60%26cos60%2660%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
The angles can be evaluated and left in the surd form.
Answer:
The pressure drop is 269.7N/m^2
Explanation:
∆P = ∆h × rho × g
∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2
∆P = 0.032×860×9.8 = 269.7N/m^2
Answer:
slenderness ratio = 147.8
buckling load = 13.62 kips
Explanation:
Given data:
outside diameter is 3.50 inc
wall thickness 0.30 inc
length of column is 14 ft
E = 10,000 ksi
moment of inertia 

Area 


r = 1.136 in
slenderness ratio 

buckling load 


Answer:
Explanation:
There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.
We are given the following information:
y₁ = 3 m
y₂ = 3 m − 2.4 m = 0.6 m
t₂ − t₁ = 0.4 s
a = -9.8 m/s²
t₀ = 0 s
v₀ = 0 m/s
We need to find y₀.
Use a constant acceleration equation:
y = y₀ + v₀ t + ½ at²
Evaluated at point 1:
3 = y₀ + (0) t₁ + ½ (-9.8) t₁²
3 = y₀ − 4.9 t₁²
Evaluated at point 2:
0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²
0.6 = y₀ − 4.9 t₂²
Solve for y₀ in the first equation and substitute into the second:
y₀ = 3 + 4.9 t₁²
0.6 = (3 + 4.9 t₁²) − 4.9 t₂²
0 = 2.4 + 4.9 (t₁² − t₂²)
We know t₂ = t₁ + 0.4:
0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)
0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))
0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)
0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)
0 = 2.4 − 3.92 t₁ − 0.784
0 = 1.616 − 3.92 t₁
t₁ = 0.412
Now we can plug this into the original equation and find y₀:
3 = y₀ − 4.9 t₁²
3 = y₀ − 4.9 (0.412)²
3 = y₀ − 0.83
y₀ = 3.83
Rounded to two significant figures, the height of the tree is 3.8 meters.
Explanation:
Population inversion is a process of achieving more electrons in the higher energy state than the lower energy state. In order to achieve population inversion, we need to supply energy to the laser medium. The process of supplying energy to the laser medium is called pumping.