Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>
<span>Solution:
A) A charge q </span>under an electric field of intensity E will experience a force F equal to:

In our problem we have
and
, so we can find the magnitude of the electric field:

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to

Therefore, the magnitude of the force acting on the proton will be

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.
I say that the answere would be B
Answer:
you
Explanation:
cause your reading this and breathing
The final position of the object after 2 s is 11 m.
Motion: This can be defined as the change in position of a body.
⇒ Formula:
- x = x₀+v₀t+1/2(at²)........................ Equation 1
⇒ Where:
- x = Final position of the object
- x₀ = Starting position
- v₀ = Starting velocity
- t = time
- a = acceleration
From the question,
⇒ Given:
- x₀ = 4.5 m/s
- t = 2 s
- x₀ = 2m
- a = 0 m/s²
⇒ Substitute these values into equation 1
- x = 2+(4.5×2)+1/2(0²×2)
- x = 2+9+0
- x = 11 m
Hence, The final position of the object after 2 s is 11 m
Learn more about motion here: brainly.com/question/15531840