Which of these is considered a contact force? *

Four +2 µC point charges are at the corners of a square of side 2 m. Find the potential at the center of the square (relative to

Romashka [77]

Answer:

(a) 51428.59 J/C

(b) 25714.29 J/C

(c) 0 J/C

Explanation:

Parameters given:

Q1 = 2 * 10^-6 C

Q2 = 2 * 10^-6 C

Q3 = 2 * 10^-6 C

Q4 = 2 * 10^-6 C

=> Q1 = Q2 = Q3 = Q4 = Q

Side of the square = 2m

The center of the square is the midpoint of the diagonals, i.e. Using Pythagoras theorem:

BD² = 2² + 2²

BD² = 8

BD = √(8) = 2.8m

OD = 1.4m

(The attached diagram explains better)

Hence, the distance between the center and each point charge, r, is 1.4m.

Electric Potential, V = kQ/r

k = Coulombs constant

(a) If all charges are positive:

V(Total) = V1 + V2 + V3 + V4

V1 = Potential due to Q1

V2 = Potential due to Q2

V3 = Potential due to Q3

V4 =Potential due to Q4

Since Q1 = Q2 = Q3 = Q4 = Q

=> V1 = V2 = V3 = V4

=> V(Total) = 4V1

V = (4 * 9 * 10^9 * 2 * 10^-6)/1.4

V = 51428.59J/C

(b) If 3 charges are positive and 1 is negative:

Since Q1 = Q2 = Q3 = Q

and Q4 = -Q

The total potential becomes:

V(Total) = V1 + V2 + V3 - V4

Since V1, V2, V3 and V4 have the same value,

V(Total) = V1 + V2

V(Total) = 2V1

V(Total) = (2 * 9 * 10^9 * 2 * 10^-6)/1.4

V(Total) = 25174.29 J/C

(c) Two charges are positive and two are negative:

Since Q1 = Q2 = Q

and Q3 = Q4 = -Q

The total potential becomes:

V(Total) = V1 + V2 - V3 - V4

Since V1, V2, V3 and V4 have the same value,

V(Total) = 0 J/C

8 0

3 years ago

Higher-frequency sound waves are perceived as having _____.

Alona [7]

Higher frequency sound waves are perceived as having higher pitch

7 0

3 years ago

What graph would best represent acceleration as a function of mass when a constant force is applied?

Alex17521 [72]

The graphs of the acceleration as a function of mass and of the acceleration as a function of force are in attachment.

Explanation:

To answer both parts of the question, we refer to Newton's second law, which states that:

$F=ma$

where

F is the net force on an object

m is the mass of the object

a is its acceleration

a)

To answer this part, we re-arrange the previous equation as follows:

$a=\frac{F}{m}$

Therefore, we notice that if a constant force is applied, the acceleration is inversely proportional to the mass of the object:

$a \propto \frac{1}{m}$

this means that if the mass increases, the acceleration decreases, and if the mass decreases, the acceleration increases.

In a graph of acceleration vs mass, the curve representing this relationship would be a hyperbole. The graph is shown as the first graph in the attached picture.

b)

As before, we re-arrange the previous equation as follows:

$a=\frac{F}{m}$

Here we notice that if the object has a constant mass, the acceleration is directly proportional to the force applied on the object:

$a \propto F$

this means that when the force increases, the acceleration increases, and when the force decreases, the acceleration decreases.

In a graph of acceleration vs force, the curve representing this relationship would be a straight line, as shown in the second graph in the attached picture.

Learn more about Newton's second law of motion here:

brainly.com/question/3820012

#LearnwithBrainly