To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as
Here,
k = Coulomb's constant
q = Charge of proton and electron
r = Distance
Replacing we have that,
The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.
The acceleration of the electron is given as
The acceleration of the proton is given as,
The change in electric potential energy is given by:
ΔU = ΔVq
ΔU = change in PE, ΔV = potential difference, q = charge
Given values:
ΔV = 50V - 20V = 30V, q = 0.14C
Plug in and solve for ΔU:
ΔU = 30(0.14)
ΔU = 4.2J
The formula to solve this is (.80)^3 X 6 and the answer would be 3.1 feet. That is how high the ball will rebound after its third bounce. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
Answer:
a = 2.84 m/s²
Explanation:
Given that,
Net force, F = 2500 N
Mass of the car, m = 880 kg
We need to find the acceleration of the car. Net force is given by :
F = ma
So, the acceleration of the car is 2.84 m/s².
Mercury a terrestrial. It isn't made of gas.
