Answer:

direction is Horizontal
Explanation:
As we know that the string is horizontal here
so the tension force in the string is due to electrostatic force on it
now we will have

so here the force is tension force on it


now we have


direction is Horizontal
Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
The magnitude of the resultant is
√ (22² + 2.2²) = √ (484 + 4.84) = √488.84 = 22.11 m/s .
The direction of the resultant is
tan⁻¹(22N / 2.2E) = tan⁻¹(10) = 5.71° east of north .
Answer:
I think the answer is b but i cannot confirm cuz this is a challenging question
The answer would be a radio wave