Answer:
The object is also positively charged because same or alike charges repel
Explanation:
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
Polarizing filters: used to show light has some properties of a wave in that way of property is that light can be thought of as traveling forward in waves, with the <span>wave. sorry if this is too late but google is good</span>
<u>Answer:</u>
Resultant displacement is 4 km north.
<u>Explanation:</u>
Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.
A runner traveled 15 kilometers north then backtracked 11 kilometers south before stopping,
So initial displacement, 15 kilometers north = 15 j km
Second displacement, 11 kilometers south = -11 j km
Total displacement = 15 j -11 j = 4 j km
Total displacement is 4 km north.
