Answer:
<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>
Explanation:
For a charged particle, the force on the particle is given as

also recall that work is force times the distance traveled
work = F x d
so, the work on the particle = F x d,
where the distance traveled by the particle in one revolution = 
Work on a particle = 2πrF = 
This work is proportional to the energy of the particle.
And the work is also proportional to the radius of travel of the particles.
Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.
Answer: c) Positive acceleration describes an increase in speed; negative acceleration describes a decrease in speed.
Hey
On Earth, you can move or you can not move. if you are moving 50 mph that means relative to Earth (not the Andromeda galaxy). When you start moving (accelerating) you are now moving relative to Earth. when you start slowing down (decelerating [most scientists just say you have negative acceleration]) you are starting to match your velocity to Earth's velocity.
Hope it helped and made any sence at all.
Spiky bob your answerer.
Answer: 4 ft/s
Explanation:
Given
height of man
speed of person 
height if street light
Let x be the distance between person and street light and y be the length of his shadow
From diagram
as the two triangle ADE and ABC are similar therefore we can say that




differentiate above Equation w.r.t time we get


T is in seconds (s)
<span>2pi is dimensionless </span>
<span>L is in meters (m) </span>
<span>g is in meters per second squared (m/s^2) </span>
<span>so you can write the equation for the period of the simple pendulum in its units... </span>
<span>s=sqrt(m/(m/s^2)) </span>
<span>simplify</span>
<span>s=sqrt(m*s^2*1/m) cancelling the m's </span>
<span>s=sqrt(s^2) </span>
<span>s=s </span>
<span>therefore the dimensions on the left side of the equation are equal to the dimensions on the right side of the equation.</span>
I don’t know sorry ;khbadkhb didhwbck( khwdicdwbihwd