A proton initially has V = (-5.60)î + (-15.0)ĵ + (4.20)k and then 2.20 s later has V = (-2.90)i + (-15.0)j + (-6.30)k (in meters per second). (a) For that 2.20 s, what is the proton's average acceleration a avg in unit vector notation, (b) in magnitude, and (c) the angle between á avg and the positive direction of the x axis?

a displacement distance is size of displ

answer///scientific method;

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m