Party hearing. as the number of people at a party increases, you must raise your voice for a listener to hear you against the background noise of the other partygoers. however, once you reach the level of yelling, the only way you can be heard is if you move closer to your listener, into the listener's "personal space." model the situation by replacing you with an isotropic point source of fixed power p and replacing your listener with a point that absorbs part of your sound waves. these points are initially separated by ri = 1.90 m. if the background noise increases by δβ = 4.90 db, the sound level at your listener must also increase. what separation rf is then required?

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)

Yup your right can give me more detail