An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5 m/s . It then flies a further distance of 46100 m , and afterwards, its velocity is 40.7 m/s . Find the airplane's acceleration.

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.