Thanks for asking your question here. I hope this response provides clarity. Feel free to ask additional questions. The moment resulting from the two forces about point O is 376 lb-ft counterclockwise.
Answer:
The flux across the cube's surface is
.
Solution:
According to the details provided:
Cube edge length, a = 8.0 cm =
.
Volume charge density,
.
Now,
To find the electric flux:

where
= electric flux
= permittivity of vacuum.
The volume charge density for this scenario is described by:

Cube volume,
.
Thus,
.
The total charge can be derived from equation (2):
.
.
Now, insert the value of 'q' into equation (1):
.
Answer:

Explanation:
The stone reaches the top of the flagpole at both t = 0.5 s and t = 4.1 s
therefore, the total duration of the upwards motion above the peak of the pole is provided as

now we have



this indicates the speed at the flagpole's top
at this point we have



the height of the flagpole is stated as



The electric force between two objects is expressed as being proportional to the product of their charges and inversely proportional to the square of the distance separating them. In this instance, the distance between the first two charges is 19 cm. We formulate the equation k q1 q3/ (x)^2 = k q2 q3/ (19-x)^2, where x denotes the separation between q1 and q3. The charge q3 cancels out, and q2 is used in absolute terms. The resulting value of x is 5.79 cm.