Answer:
The acceleration of the proton is 2.823 x 10¹⁷ m/s²
The acceleration of the electron is 5.175 x 10²⁰ m/s²
Explanation:
Given;
distance between the electron and proton, r = 7 x 10⁻¹⁰ m
mass of proton, = 1.67 x 10⁻²⁷ kg
mass of electron, = 9.11 x 10⁻³¹ kg
The attractive force between the two charges is given by Coulomb's law;
where;
k is Coulomb's constant = 9 x 10⁹ Nm²/c²
Acceleration of proton is given by;
F = ma
Acceleration of the electron is given by;
Answer:
Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.
Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.
Explanation:
Part a
When Road is Level
The stopping sight distance is given as
Here
Substituting values
So the minimum stopping sight distance is 563.36 ft.
Part b
When Road has a maximum grade of 4%
The stopping sight distance is given as
Here
For upgrade of 4%, Substituting values
<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>
For downgrade of 4%, Substituting values
<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>
As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.