Answer:
a.Attractive

Explanation:
When it comes to charges, the charges which are alike repel each other and the charges which are different will attract each other.
Here, there is a proton and electron which are different particles hence, they will attract each other.
= Charge of electron and proton = 
r = Distance between them = 997 nm
k = Coulomb constant = 
Force is given by

The force of attraction between the particles will be 
The correct answer for the first one is A) Breaking rocks into smaller pieces. The correct answer to the second one is A) It has more energy. Hope this helps.
Answer:
Explanation:
You can approach an expression for the instantaneous velocity at any point on the path by taking the limit as the time interval gets smaller and smaller. Such a limiting process is called a derivative and the instantaneous velocity can be defined as.#3
For the special case of straight line motion in the x direction, the average velocity takes the form: If the beginning and ending velocities for this motion are known, and the acceleration is constant, the average velocity can also be expressed as For this special case, these expressions give the same result. Example for non-constant acceleration#1
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake