Which of the following questions would be answered using concepts from physics?

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

The frequency of vibrations of a vibrating violin string is given by f = 1 2l t ρ where l is the length of the string, t is its tension, and ρ is its linear density.† (a) find the rate of change of the frequency with respect to the following. (i) the length (when t and ρ are constant) (ii) the tension (when l and ρ are constant) (iii) the linear density (when l and t are constant) (b) the pitch of a note (how high or low the note sounds) is determined by the frequency f. (the higher the frequency, the higher the pitch.) use the signs of the derivatives in part (a) to determine what happens to the pitch of a note for the following. (i) when the effective length of a string is decreased by placing a finger on the string so a shorter portion of the string vibrates df dl 0 and l is ⇒ f is ⇒ (ii) when the tension is increased by turning a tuning peg df dt 0 and t is ⇒ f is ⇒ (iii) when the linear density is increased by switching to another string df dρ 0 and ρ is ⇒ f is ⇒

Carbon is allowed to diffuse through a steel plate 15 mm thick. the concentrations of carbon at the two faces are 0.65 and 0.30 kg c/m^3 fe, which are maintained constant. if the preexponential and activation energy are 6.2 x 10-7 m2 /s and 80,000 j/mol, respectively, compute the temperature at which the diffusion flux is 1.43 x 10^-9 kg/m^2 -s.