Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uniform electric field. The electric field exerts a constant hor-izontal force on a charged object. A 15 g charged projectile is launched with a speed of 6.0 m/s at an angle 35° above the hori-zontal. It lands 2.9 m in front of the launcher. What is the magni-tude of the electric force on the projectile?

Magnitude of electric force = 0.03345 N

Explanation:

We are given;

Mass; m = 15g = 0.015kg

Angle above horizontal; θ = 35°

Speed; v = 6 m/s

Horizontal displacement; d = 2.9m

Now formula for time of flight is given as;

time of flight; t = (2Vsinθ)/g

Thus, plugging in values, we have

t = (2 x 6.0 x sin35)/9.8

t = (12 x 0.5736)/9.8

t = 0.7024 s

Now, let's find the acceleration

The formula for horizontal displacement is given by;

d = (Vcosθ)t + (1/2)at²

Plugging in the relevant values ;

2.9 = [6(cos35) x 0.7024] + (1/2)a(0.7024)²

2.9 = (4.2144 x 0.8192) + (0.2467)a

2.9 = 3.45 + (0.2467)a

(0.2467)a = 2.9 - 3.45

a = -0.55/0.2467

a = -2.23 m/s²

Since we are looking for the magnitude of the electric force, we will take the absolute value of a. Thus, a = 2.23 m/s²

We know that F = ma

Thus,Force = 0.015kg x 2.23m/s² =

= 0.03345 N