The magnitude of force acting on wire will be 1.90 N and the direction of the force acting on the wire will be 41.9 degrees below the negative y axis.
Explanation:
It is known that the force acting on a current carrying conductor placed in a magnetic field is

Here B is the magnetic field, I is the current flowing through the wire and L is the length of the wire which is given as 44 cm.
Since the wire is bended in the middle at right angle so the length of the two sides of the wire will be 22 cm each. Also one part is lying over z axis and another part lies in the plane of xy in the equation of line y = 2x. So the slope of this wire will be

This will be equal to tan θ.
So θ = tan⁻¹ (2) =63.4°
Then, the length of the wire will be written as components of i, j and k.


Then,
F = I (L × B)

![F = 20.5 (\left[\begin{array}{ccc}i&j&k\\0.098&0.197&-0.22\\0.316&0&0\end{array}\right] )](https://tex.z-dn.net/?f=F%20%3D%2020.5%20%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0.098%260.197%26-0.22%5C%5C0.316%260%260%5Cend%7Barray%7D%5Cright%5D%20%29)


The magnitude of force on the wire will be

And the direction can be found by the tan inverse of the ratio of k component to j component of the force.

So the magnitude of force acting on wire will be 1.90 N and the direction of the force acting on the wire will be 41.9 degrees below the negative y axis.
Answer:
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
Explanation:
Given :
Frequencies of the sinusoids,
, and

Sampling rate 
The positive frequencies at the output of the sampling system are :

When n = 0,

when n = 1,



When n = 2,


Therefore, the first six positive frequencies present in the replicated spectrum are :
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
<em>Class I .</em> . .
The fulcrum is between the effort and the load.
The mechanical advantage may be any positive number, more or less than ' 1 '.
<em>Class II .</em> . .
The load is between the fulcrum and the effort.
The mechanical advantage is always greater than ' 1 '.
<em>Class III .</em> . .
The effort is between the fulcrum and the load.
The mechanical advantage is always less than ' 1 '.
The car travels a distance of
(8.0 m/s) (5 s) + 1/2 (10.0 m/s²) (5 s)² = 165 m
This type of energy is called non mechanical energy so the answer is C