Answer:
D) This is the correct answer
Explanation:
In this exercise the two ball loads are suspended by a thread.
To answer this exercise, let us remember that charges of the same sign repel and charges of a different sign attract.
Therefore, for the system to maintain equilibrium, the two charges must be of the same sign.
When examining the different proposals
A) in this case, as a sphere has no charge, there is no electric force and the induced charge is of the opposite sign, so the spheres attract each other
B) in this case there is an electric force, but being of a different sign, the force is attractive so the system is not in equilibrium
C) as the charges are of different magnitude the system does not have equal angles
D) This is the correct answer, since the charges have the same magnitude and are of the same sign, so the force is repulsive and is counteracted by the weight component
F_e = W sin θ
Answer:
the standard way the body is positioned when using anatomical terminology ... invisible line that runs vertically through the center of the axial region.
Explanation:
Answer:
19 and below because a whole number is just a number that isn't a fraction. They are basic counting numbers
Explanation:
Answer:
w = 11.211 KN/m
Explanation:
Given:
diameter, d = 50 mm
F.S = 2
L = 3
Due to symmetry, we have:
![Ay = By = \frac{w * 6}{2} = 3w](https://tex.z-dn.net/?f=%20Ay%20%3D%20By%20%3D%20%5Cfrac%7Bw%20%2A%206%7D%7B2%7D%20%3D%203w%20)
![P_c_r = 3w * F.S = 3w * 2.0 = 6w](https://tex.z-dn.net/?f=%20P_c_r%20%3D%203w%20%2A%20F.S%20%3D%203w%20%2A%202.0%20%3D%206w%20)
![I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7](https://tex.z-dn.net/?f=%20I%20%3D%20%5Cfrac%7B%5Cpi%7D%7B64%7D%20%280.05%29%5E4%20%3D%203.067%2A10%5E-%5E7%20)
To find the maximum intensity, w, let's take the Pcr formula, we have:
![P_c_r = \frac{\pi^2 E I}{(KL)^2}](https://tex.z-dn.net/?f=%20P_c_r%20%3D%20%5Cfrac%7B%5Cpi%5E2%20E%20I%7D%7B%28KL%29%5E2%7D%20)
Let's take k = 1
Substituting figures, we have:
![6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}](https://tex.z-dn.net/?f=%206w%20%3D%20%5Cfrac%7B%5Cpi%5E2%20%2A%20200%2A10%5E9%20%2A%203.067%2A10%5E-%5E7%7D%7B%281%20%2A%203%29%5E2%7D%20)
Solving for w, we have:
![w = \frac{67266.84}{6}](https://tex.z-dn.net/?f=%20w%20%3D%20%5Cfrac%7B67266.84%7D%7B6%7D%20)
w = 11211.14 N/m = 11.211 KN/m
Since Area, A= pi * (0.05)²
. This means it is safe
The maximum intensity w = 11.211KN/m