A particle executes simple harmonic motion
with an amplitude of 3.43 cm.
At what positive displacement from the
midpoint of its motion does its speed equal
one half of its maximum speed?
Answer in units of cm.

The positive displacement x = 0.27 cm.

Explanation:

The formula to calculate displacement of a particle executing simple harmonic motion is given by,

                                   x  =  A s i n ( ω t )

where x represents the displacement,

          A represents the amplitude,

          ω represents the angular frequency,

           t represents the time.

Given, amplitude A = 3.43 cm,   t = 1 / 2  = 0.5 s

                                  x = A s i n ( ω t )

Since ω = 2πf,

                                  x = 3.43 sin (23.14500.5)

                                     = 3.43 sin(157)

                                  x = 0.27 cm.

answer: i don't know the answer good luck

explanation:

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop