Answer:
![M = \left[\begin{array}{ccc}cos \ 60&0\\0&-sin \ 60\end{array}\right]](https://tex.z-dn.net/?f=M%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%20%5C%2060%260%5C%5C0%26-sin%20%5C%2060%5Cend%7Barray%7D%5Cright%5D)
Explanation:
To find the matrix, let's decompose the vectors, the rotated angle is (-60C) for the prime system
x ’= x cos (-60)
y ’= y sin (-60)
we use
cos 60 = cos (-60)
sin 60 = - sin (-60)
we substitute
x ’= x cos 60
y ’= - y sin 60
the transformation system is
x '
the transformation matrix is
![M = \left[\begin{array}{ccc}cos \ 60&0\\0&-sin \ 60\end{array}\right]](https://tex.z-dn.net/?f=M%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%20%5C%2060%260%5C%5C0%26-sin%20%5C%2060%5Cend%7Barray%7D%5Cright%5D)
The magnetic field at the center of the circular loop is 3.24 x 10⁻⁴ T.
<h3>Current in the wire</h3>
The current in the wire is calculated as follows;
V = IR
I = V/R
I = 16.7/0.251
I = 66.53 A
<h3>Magnetic field at the center of the loop</h3>
The magnetic field at the center of the circular loop is calculated as follows;

Learn more about magnetic field here: brainly.com/question/7802337
Answer:
The mass of the block, M =T/(3a +g) Kg
Explanation:
Given,
The upward acceleration of the block a = 3a
The constant force acting on the block, F₀ = Ma = 3Ma
The mass of the block, M = ?
In an Atwood's machine, the upward force of the block is given by the relation
Ma = T - Mg
M x 3a = T - Ma
3Ma + Mg = T
M = T/(3a +g) Kg
Where 'T' is the tension of the string.
Hence, the mass of the block in Atwood's machine is, M = T/(3a +g) Kg
Answer: D. 25 m/s Downward
Explanation:
V = (g)(t)
V = (9.8 m/s²)(2.5)
V = 24.5 m/s