In the final part of its life cycle a particular rotating spherical star expands to four times its normal diameter. assume the mass remains constant, the mass is uniformly distributed before the expansion and the mass is uniformly distributed after the expansion. how is the period of rotation affected? the period decreases to one quarter its normal period. the period increases by a factor of four. the period decreases to one eighth its normal period. the period increases by a factor of two. the period increases by a factor of sixteen. the period decreases to one sixteenth its normal period. the period is unchanged. the period increases by a factor of eight. the period decreases to one half its normal period.

Answer8.0m;

Ithink the answer is fusion 

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: where m is the mass of the arrow and v its velocity.  all the other options, instead, do not increase the kinetic energy of the system: -  pulling farther back on the string : this will just increase the elastic potential energy of the system-  holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)-  decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)