Answer:
The resulting magnetic force on the wire is -1.2kN
Explanation:
The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is
F = I (L*B)
Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)
L x B = ![\left[\begin{array}{ccc}i&j&k\\2&0&0\\30&-40&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2%260%260%5C%5C30%26-40%260%5Cend%7Barray%7D%5Cright%5D)
= (0, 0, -80)
we can now solve
F = I (L x B) = I (-80)
F = -1200 kmN
F = -1200 kN * 10⁻³
F = -1.2kN
To solve the problem it is necessary to identify the equation in the manner given above.
This equation corresponds to the displacement of a body under the principle of simple harmonic movement.
Where,
PART A) Our equation corresponds to
Therefore the value of omega is equivalent to that of
From the definition we know that the period as a function of angular velocity is equivalent to
This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the 
Is equivalent to . So the maximum speed that the body can reach is,
Therefore the maximum felocity will be 5ft / s
PART B) The period of graph is the time taken to reach from one maximum point to next point maximum point, then
Answer:
Hello there Dude answer is B :D hope it helped mark me brainliest.
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