Answer:
T1 = 131.4 [N]
T2 = 261 [N]
Explanation:
To solve this problem we must make a sketch of how will be the semicircle, for this reason we conducted an internet search, to find the scheme of the problem. This scheme is attached in the first image.
Then we make a free body diagram, with this free body diagram, we raise the forces that act on the body. Since it is a problem involving static equilibrium, the sum of forces in any direction and moments must be equal to zero.
By performing a sum of forces on the Y axis equal to zero we can find an equation that relates the forces of tension T1 & T2.
The second equation can be determined by summing moments equal to zero, around the point of application of the T1 force. In this way we find the T2 force.
The value of T2, is replaced in the first equation and we can find the value for T1.
Therefore
T1 = 131.4 [N]
T2 = 261 [N]
The free body diagram and the developed equations can be seen in the second attached image.
Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?
Answer:
17.32 m
Explanation:
From coulomb's Law,
F = kqq'/r²........................... Equation 1
Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.
make r the subject of the equation above
r = √(kqq'/F)..................... Equation 2
From the question,
Given: q = q' = 0.005 C, F = 750 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute these values into equation 2
r = √(9.0×10⁹×0.005×0.005/750)
r = √(300)
r = 17.32 m.
Hence the distance between the pair of charges = 17.32 m
They all compact like are close together
but the points are only 25
Answer:
0.0768 revolutions per day
Explanation:
R = Radius
= Angular velocity
As the mass is conserved the angular momentum is conserved

Moment of intertia for solid sphere

Moment of intertia for hollow sphere

Dividing the moment of inertia

From the first equation

The angular velocity, in revolutions per day, of the expanding supernova shell is 0.0768 revolutions per day