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:
The angles can be evaluated and left in the surd form.
Answer:
When a pilot pushes the top of the right pedal, it activates the brakes on the right main wheel/wheels, and when the pilot pushes the top of the left rudder pedal, it activates the brake on the left main wheel/wheels. The brakes work in a rather simple way: they convert the kinetic energy of motion into heat energy.
Answer:
The inductance of the inductor is 0.051H
Explanation:
From Ohm's law;
V = IR .................. 1
The inductor has its internal resistance referred to as the inductive reactance, X
, which is the resistance to the flow of current through the inductor.
From equation 1;
V = IX
X = 
................ 2
Given that; V = 240V, f = 50Hz, 
= 
, I = 15A, so that;
From equation 2,
X= 
= 16Ω
To determine the inductance of the inductor,
X = 2fL
L = 
= 
= 0.05091
The inductance of the inductor is 0.051H.
Answer:
<h3><em>Transistor switches can be used to switch a low voltage DC device (e.g. LED’s) ON or OFF by using a transistor in its saturated or cut-off </em><em>state</em></h3>
- <em>. Cut-off </em><em>Region</em>
<em>Here the operating conditions of the transistor are zero input base current ( IB ), zero output collector current ( IC ) and maximum collector voltage ( VCE ) which results in a large depletion layer and no current flowing through the device. Therefore the transistor is switched “Fully-OFF”.</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>2</em><em>.</em><em>Saturation </em><em>Region</em>
<em>Here the transistor will be biased so that the maximum amount of base current is applied, resulting in maximum collector current resulting in the minimum collector emitter voltage drop which results in the depletion layer being as small as possible and maximum current flowing through the transistor. Therefore the transistor is switched “Fully-ON”.</em>
Explanation:
hope it helps
