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?
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Ответ:
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
Ответ:
Translation means it just moves the shape, it does not rotate or reflect the shape in any way.