Answer:
Q = 4 Q₀
Explanation:
This is an exercise on capacitors, where the capacitance is
C = 
if we apply the given conditions
C = \epsilon_{o} \ \frac{2A}{0.5d}
C = 4 \epsilon_{o} \ \frac{A}{d}
let's call the capacitance Co with the initial values
C₀ = \epsilon_{o} \ \frac{A}{d}
C = 4 C₀
The charge on each plate of a capacitor is
Q = C ΔV
If the potential difference is maintained, the new charge is
Q = 4 C₀ ΔV
let's call
Q₀ = C₀ ΔV
we substitute
Q = 4 Q₀
Answer: 10 m/s
We're told the speed is constant, so it's not changing throughout the time period given to us. So throughout the entire interval, the speed is 10 m/s.
Answer:
-10.8°, or 10.8° below the +x axis
Explanation:
The x component of the resultant vector is:
x = 3.14 cos(30.0°) + 2.71 cos(-60.0°)
x = 4.07
The y component of the resultant vector is:
y = 3.14 sin(30.0°) + 2.71 sin(-60.0°)
y = -0.777
Therefore, the angle between the resultant vector and the +x axis is:
θ = atan(y / x)
θ = atan(-0.777 / 4.07)
θ = -10.8°
The angle is -10.8°, or 10.8° below the +x axis.
Answer:
The positively charged ball moves between both charged plates till the plates and the ball all become neutral.
Check Explanation for more.
Explanation:
Let the ball be in square brackets, and the plates in normal brackets.
(+) [+] (-)
From the law that like charges repel and unlike charges attract.
The positive ball would go first to the negatively charged plate. After which, the ball would hold more negative charges overall than before.
Because the ball is now more negatively charged, it then travels towards the positive plate. In the same manner, the ball would transfer negative electrons to the positive plate.
So, when leaving the positive plate, the ball would be more positive and be drawn towards the negative plate once more. In doing so, it would make the negative plate more positive.
Then, the ball again holds more negative electrons and is drawn towards the positive plate once more.
This back and forth process continues until the once-positive and once-negative plates become neutral, that is, they are discharged.
The ball hanging on the insulated thread becomes neutral too at this point.
Hope this Helps!!!
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.
