A proton enters a region filled with a magnetic field B = B0k. The proton enters with a velocity v = 103
i + 2 × 102k m s−1
. Assume that i points toward the right of the page and k points toward the
top.
(a) Sketch the setup, show the relevant vectors/forces and the direction of the orbit of the proton.
(b) What is the frequency of the proton in its orbit? Or, how long does it take the proton to make
one orbit?
(c) The proton travels 10 cm along the field, how many turns does it make?

Iwould say the answer is d

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)

You need to calculate the string in between two masses