Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each travels before its speed is reduced by half is measured. one tire is inflated to a pressure of 40 psi and goes a distance of 18.8 m ; the other is at 105 psi and goes a distance of 92.7 m . assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g = 9.80 m/s2 .

The horse latitudes-high-pressure zones about 30 degrees north and 30 degrees south of the equator.-warm air traveling away from the equator cools and sinks-weather tends to be clear and drythe doldrums-low-pressure near equator-air rises and spreads out toward the poles-rising moist air produces clouds and heavy rainthe westerlies-blow from the west-move from the horse latitudes toward the poles.-bring storms across much of the united statesthe tradewinds-blow from the east-move from the horse latitudes toward the equator  -strong, steady winds

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)

I’m not sure about this question