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.
Answer:
B
Explanation:
Adding a dopant is correct on edge.
Answer:
The center of mass of the two-ball system is 7.05 m above ground.
Explanation:
<u>Motion of 0.50 kg ball:</u>
Initial speed, u = 0 m/s
Time = 2 s
Acceleration = 9.81 m/s²
Initial height = 25 m
Substituting in equation s = ut + 0.5 at²
s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m
Height above ground = 25 - 19.62 = 5.38 m
<u>Motion of 0.25 kg ball:</u>
Initial speed, u = 15 m/s
Time = 2 s
Acceleration = -9.81 m/s²
Substituting in equation s = ut + 0.5 at²
s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m
Height above ground = 10.38 m
We have equation for center of gravity

m₁ = 0.50 kg
x₁ = 5.38 m
m₂ = 0.25 kg
x₂ = 10.38 m
Substituting

The center of mass of the two-ball system is 7.05 m above ground.
Explanation:
The object is moving along the parabola y = x² and is at the point (√2, 2). Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.
First, we need to find the radius of curvature. This is given by the equation:
R = [1 + (y')²]^(³/₂) / |y"|
y' = 2x and y" = 2:
R = [1 + (2x)²]^(³/₂) / |2|
R = (1 + 4x²)^(³/₂) / 2
At x = √2:
R = (1 + 4(√2)²)^(³/₂) / 2
R = (9)^(³/₂) / 2
R = 27 / 2
R = 13.5
So the centripetal force is:
F = m v² / r
F = m (5)² / 13.5
F = 1.85 m