In order to change a 2.15 kg handcart's velocity from 1.2 m/s east to 1.8 m/s east, what's the magnitude of the impulse that
should be applied to the handcart?
a
38.7 N-s
b 12.9 N-S
2.58 N-S
d 3.87 N-s
e
1.29 N. S

Six solutions are made, each with a y concentration of 1.0 μm and varying concentrations of ab as shown below. based on the concentrations, rank the solutions in decreasing order of reaction rate.

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 x-coordinate of a particle in curvilinear motion is given by x = 3.1t3 - 4.9t where x is in feet and t is in seconds. the y-component of acceleration in feet per second squared is given by ay = 2.3t. if the particle has y-components y = 0 and vy = 5.0 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 1.8 sec. sketch the path for the first 1.8 seconds of motion, and show the velocity and acceleration vectors for t = 1.8 sec. answers: v = ft/sec a = ft/sec2