Answer:
W = 2.74 J
Explanation:
The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.
This is the electrostatic equivalent of the work-energy theorem.
where the potential energy is defined as follows
Let's first calculate the distance 'r' for both positions.
Now, we can calculate the potential energies for both positions.
Finally, the total work done on the moving particle can be calculated.
Answer:
<h2>A. 180 miles</h2><h2>B. 60 miles</h2><h2 />
Explanation:
In this problem, we are required to solve for the total distance that the car travelled. and the displacement
A) the distance travelled by car
this can be gotten by summing all the distances the car has travelled.
i,e total distance= 60 miles+120 miles
total distance= 180 miles
B) the displacement of the car
the displacement can be gotten by subtracting the final distance from the initial distance
final distance = 120 miles
initial distance= 60 miles
displacement= 120-60= 60 miles
IM sure there is C, D, and E in kuiper belts, but not really sure of silicon and iron
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:
m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,
m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1
The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is
(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2
Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
