Answer:
2.4 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.08 kg)(0.5 m/s) + (0.05 kg)(0 m/s) = (0.08 kg)(-0.1 m/s) + (0.05 kg) v
0.04 kg m/s = -0.08 kg m/s + (0.05 kg) v
0.12 kg m/s = (0.05 kg) v
v = 2.4 m/s
Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ = 
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| = 
(|r₂₁|)² = 148.25

= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Answer:
A super conductor is a perfect conductor that has zero resistance. It doesn't just have very low resistance and conducts electricity well, it has ZERO resistance and conducts electricity perfectly with no losses at all
There are two ways to solve this. The longer way is to use those equations to calculate numbers for total distance.
The easier way is to find the area under the graph. That's right, AREA UNDER VELOCITY-TIME graph is the TOTAL DISTANCE travelled!
it's a shortcut.
Let's split up the area into a triangle and rectangle:
Triangle = 0.5(4-0)(10-0) = 20 m
Rectangle = (6-4)(10-0) = 20 m
Total distance = 40 m!
Well, if a charger conductor is touched to another object or close enough to touching the object then the conductor can transfer its charge to that object. Conductors allow for electrons to be transported from particle to particle, so a charged object will always distribute its charge until the repulsive forces are minimized.