The answer for this question is acceleration
Answer:
The value of the distance is
.
Explanation:
The velocity of a particle(v) executing SHM is
![v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)](https://tex.z-dn.net/?f=v%20%3D%20%5Comega%20%5Csqrt%7BA%5E%7B2%7D%20-%20x%5E%7B2%7D%7D~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%60~%281%29)
where,
is the angular frequency,
is the amplitude of the oscillation and
is the displacement of the particle at any instant of time.
The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e.,
.
The maximum velocity(
) is
![v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)](https://tex.z-dn.net/?f=v_%7Bm%7D%20%3D%20%5Comega%20A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%282%29)
Divide equation (1) by equation(2).
![\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)](https://tex.z-dn.net/?f=%5Cdfrac%7Bv%7D%7Bv_%7Bm%7D%7D%20%3D%20%5Cdfrac%7B%5Csqrt%7BA%5E%7B2%7D%20-%20x%5E%7B2%7D%7D%7D%7BA%7D~~~~~~~~~~~~~~~~~~~~~~~~~~~%283%29)
Given,
and
. Substitute these values in equation (3).
![&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm](https://tex.z-dn.net/?f=%26%26%20%5Cdfrac%7B1%7D%7B4%7D%20%3D%20%5Cdfrac%7B%5Csqrt%7B15%5E%7B2%7D%20-%20x%5E%7B2%7D%7D%7D%7B15%7D%5C%5C%26or%2C%26%20A%20%3D%2014.52~cm)
I believe the answer here is <span>C).seedless, seed (nonflowering), and seed (flowering).</span><span>
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THE ANSWER IS 16 ohms or however its spelled