A piece of clay with a mass of 350 g is dropped to the floor from a height of 1.5 m (A). Calculate the KE and PE of the clay at

a. (A = 1.5 m).
b. (B = 70 cm), and
c. (C = 0 m, just before hitting the ground). d. Make an energy graph at each point​

Did the proton move into a region of higher potential or lower potential? did the proton move into a region of higher potential or lower potential? because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of higher potential to a region of lower potential.because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of lower potential to a region of higher potential.because the proton is a positive charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential.because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential.request answerpart bwhat was the potential difference that stopped the proton? express your answer with the appropriate units.î”v î” v = nothingnothingrequest answerpart cwhat was the initial kinetic energy of the proton, in electron volts? express your answer in electron volts.ki k i = nothing ev request answerprovide feedback

When is a current produced? when the terminals of an electrochemical cell are connected by a wire if the electric circuit is opened in an electrochemical cell if the electrolyte is removed from an electrochemical cell when the electrodes are reversed in an electrochemical cell

Ablock of mass m slides on a horizontal frictionless table with an initial speed v0 . it then compresses a spring of force constant k and is brought to rest. the acceleration of gravity is 9.8 m/s2. how much is the spring compressed x from its natural length? 1) x = v0*sqrt(k/(mg)) 2) x=v0*sqrt(m/k) 3) x=v0*((mk)/g) 4) x=v0*sqrt(k/m) 5) x=v0*(m/kg) 6) x=v0*sqrt((mg)/k) 7) x=(v0)^2/(2g) 8) x=v0*(k/(mg)) 9) x=(v0)^2/(2m) 10) x=v0*((mg)/k)