Formula:
F = ma
F: force (N) m: mass (kg) a: acceleration (m/s^2)
Solution:
F = ma
F = 20 × 10
= 200N
Answer:
False
Explanation:
The formula of force that exists between two charges is expressed as;
F = kq1q2/r²
If two charges separated by one meter exert a 9 N force on each other, the;
9 = kq1q2/1²
9 = kq1q2 ..... 1
If the charges are pushed to a 3 meter separation, then;
F = kq1q2/3²
F = kq1q2/9 .... 2
Divide both equations;
9/F = (kq1q2)/ kq1q2/9
9/F = kq1q2 * 9/ kq1q2
9/F = 9
F = 9/9
F = 1N
Hence if the charges are pushed to a 3 meter separation, then the force on EACH charge will be 1N. Hence the answer is False
To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.
From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

Where,
Final velocity
Initial Velocity
a = Acceleration
x = Displacement
Acceleration can be expressed in terms of the drag coefficient by means of
Frictional Force
Force by Newton's second Law
Where,
m = mass
a= acceleration
Kinetic frictional coefficient
g = Gravity
Equating both equation we have that



Therefore,


Re-arrange to find x,

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.
By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
=>3.5 - 2.75 = v1-1.5
<span>
=>v¹ = 2.25 m/s (East) </span>