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 
Answer:
Average force = 67 mn
Explanation:
Given:
Initial velocity u = 0 m/s
Final velocity v = 67 m/s
Time t = 1 ms = 0.001 sec.
Computation:
Using Momentum theory
Change in momentum = F × Δt
(v-u)/t = F × Δt
F × 0.001 = (67 - 0)/0.001
F= 67,000,000
Average force = 67 mn
Answer:
Friction of the road on the motorcycle in the opposite direction
Explanation:
Khanacademy
Answer:
B. Containing charged regions
Explanation:
The term i.e. intermolecular forces would be used to explain the attraction forces. Here the interaction would be done between molecules etc that acts between the acts & the other types of particles i.e. neighboring like atoms or ions
So in the given case, the option b would be contributed to the molecules that have intermolecular forces
hence, the option b is correct
Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,

Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?

a = -1.47 m/
(a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/
(since same friction force is applied)

s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m