Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;

Where;
d is the separation or distance between the two parallel plates;

Therefore, the necessary separation between the two parallel plates is 0.104 mm
5m/s
100m
Explanation:
Average speed is sum of distance distance traveled in a given time by a body.
Average speed= 
Distance = 20m
time = 4s
Average speed =
= 5m/s
For the spaceship;
Distance covered = speed x time
Speed = 50m/s
time = 2s
Distance covered = 50 x 2 = 100m
learn more:
Average speed brainly.com/question/5063905
#learnwithBrainly
Answer:
1.5 m
Explanation:
Let the distance from the box to the pivot be c.
Let the distance from the pivot to the effort be y.
From the question given above, the following data were obtained:
Effort force (Fₑ) = 7 N
Force of resistance (Fᵣ) = 14 N
Distance from the box to the pivot (c) = 0.75 m
Distance from the pivot to the effort (y) =?
Clockwise moment = Fₑ × y
Anticlock wise moment = Fᵣ × c
Clockwise moment = Anticlock wise moment
Fₑ × y = Fᵣ × c
7 × y = 14 × 0.75
7 × y = 10.5
Divide both side by 7
y = 10.5 / 7
y = 1.5 m
Therefore, the distance from the pivot to the effort is 1.5 m
Answer:
the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:
This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is
fr- w || = 0
The expression for friction force is that it is proportional to the coefficient of friction by normal.
fr = μ N
When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.
In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.
μ kinetic <μ static
In all this movement the normal with changed that the angle of the table remains fixed.
Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction