Answer:
The net Electric field at the mid point is 289.19 N/C
Given:
Q = + 71 nC =
Q' = + 42 nC =
Separation distance, d = 1.9 m
Solution:
To find the magnitude of electric field at the mid point,
Electric field at the mid-point due to charge Q is given by:
Now,
Electric field at the mid-point due to charge Q' is given by:
Now,
The net Electric field is given by:
The horizontal range of the missile is b) 176 m
Explanation:
The motion of the missile is a projectile motion, so it consists of two independent motions:
- A uniform motion with constant velocity along the horizontal direction
- A uniformly accelerated motion with constant acceleration (equal to the acceleration of gravity) in the vertical-downward direction
To find the time of flight of the missile, we study the vertical motion. We can use the following suvat equation:
where:
s = 75 m is the vertical displacement of the missile (the height of the building)
is the initial vertical velocity (the missile is thrown horizontally)
t is the time of flight
is the acceleration of gravity
Solving for t, we find the time of flight:
This means that the missile takes 3.91 s to reach the ground.
Now we study the horizontal motion: the missile moves with a constant horizontal velocity of
Therefore, the distance covered in a time t is
and by substituting t = 3.91 s, we find the horizontal range of the missile:
Learn more about projectile motion:
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