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:
1. Either larger or smaller than the displacement of either wave acting alone, depending on the signs of the displacements of the two waves.
Answer:
= 4.3 × 10 ⁻¹⁴ m
Explanation:
The alpha particle will be deflected when its kinetic energy is equal to the potential energy
Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C
Kinetic energy of the alpha particle = 5.28 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV)
= 8.459 × 10⁻¹³
k electrostatic force constant = 9 × 10⁹ N.m²/c²
Kinetic energy = potential energy = k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus
r = ( 9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 8.459 × 10⁻¹³
= 4.3 × 10 ⁻¹⁴ m
Pair production<span> is a direct conversion of radiant energy to matter. It is one of the principal ways in which high-energy gamma rays are absorbed in matter. </span>
