The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case,
kQ1/(r1)^2 = kQ2/(r2)^2 r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2 kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 = 1/4
The motion of a falling whirligig is different to that of a falling paper ball due to spinning.
<h3>Type of motion performed by whirligig and falling paper ball </h3>
The motion of a falling whirligig is different from the motion of a falling paper ball because the paper ball falls on the ground without spinning while on the other hand, the whirligig falls on the ground along with spinning.
The falling whirligig performs two motion i.e. one is falling on the ground and the other is spinning during motion whereas paper ball performs one motion i.e. motion in the air towards the ground so we can conclude that the motion of a falling whirligig is different than of a falling paper ball.
Learn more about motion here: brainly.com/question/453639
Does it have to be that exact word. cause it is just another term for psuedopodium
Answer:
speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s
initial kinetic energy = final kinetic energy
Explanation:
Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision
So we will have
momentum conservation along x direction
now plug in all values in it
so we have
similarly in Y direction we have
now plug in all values in it
so we have
now from 1st equation we have
so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s
Also we know that since this is an elastic collision so here kinetic energy is always conserved to
initial kinetic energy = final kinetic energy
