Answer:
Vf = 3.67 [m/s]
Explanation:
To solve this problem we must use the following equation of kinematics.

where:
Vf = final velocity [m/s]
Vi = initial velocity = 4.3 [m/s]
a = acceleration or desacceleration = 0.5 [m/s²]
x = distance = 5 [m]
Note: The negative sign in the above equation means that the velocity of the ball is decreasing (desacceleration).
Now replacing:
Vf² = (4.3)² - (2*0.5*5)
Vf² = 18.49 - 5
Vf² = 13.49
using the square root, we have.
Vf = 3.67 [m/s]
Answer:
i) C decreases
ii) Q remains constant
iii) E remains constant
iv) ΔV increases
Explanation:
i)
We know, capacitance is given by:


<em>In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.</em>
ii)
As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.
iii)
Electric field between the plates is given as:

where:
charge density, 
<em>As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.</em>
iv)
- From the basic definition of voltage we know that it is the work done per unit charge to move it through a distance.
- Here we increase the distance so the work done per unit charge increases.
Answer:
Distance 5 km, Displacement 3 km east
Explanation: