Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
The moment (torque) is given by the following equation:
![\vec{\tau} = \vec{r} \times \vec{F}\\\vec{r} \times \vec{F} = \left[\begin{array}{ccc}\^{i}&\^j&\^k\\r_x&r_y&r_z\\F_x&F_y&F_z\end{array}\right] = \left[\begin{array}{ccc}\^{i}&\^j&\3k\\0.23&0.04&0\\150&260&0\end{array}\right] = \^k((0.23*260) - (0.04*150)) = \^k (53.8~Nm)](https://tex.z-dn.net/?f=%5Cvec%7B%5Ctau%7D%20%3D%20%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7BF%7D%5C%5C%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7BF%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5C%5E%7Bi%7D%26%5C%5Ej%26%5C%5Ek%5C%5Cr_x%26r_y%26r_z%5C%5CF_x%26F_y%26F_z%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5C%5E%7Bi%7D%26%5C%5Ej%26%5C3k%5C%5C0.23%260.04%260%5C%5C150%26260%260%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5C%5Ek%28%280.23%2A260%29%20-%20%280.04%2A150%29%29%20%3D%20%5C%5Ek%20%2853.8~Nm%29)
Explanation:
The cross-product between the distance and the force can be calculated using the method of determinant. Since the z-components are zero, it is easy to calculate.
Answer:
162N
Explanation: resolving into components and equating both vertical components equal to 162 as explained in attachment
Answer:
a) E = ρ / e0
b) E = ρ*a / (e0 * r)
c) E = 0
Explanation:
Because of the geometry, the electric field lines will all have a radial direction.
Using Gauss law

Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.
We can make this cylinder of different radii to evaluate the electric field at different points.
Then:
A = 2*π*r (area of cylinder per unit of length)
Q/e0 = 2*π*r*E
E = Q / (2*π*e0*r)
Where Q is the charge contained inside the cylinder.
Inside the cable core:
There is a uniform charge density ρ
Q(r) = ρ * 2*π*r
Then
E = ρ * 2*π*r / (2*π*e0*r)
E = ρ / e0 (electric field is constant inside the charged cylinder.
Between ther inner cilinder and the tube:
Q = ρ * 2*π*a
E = ρ * 2*π*a / (2*π*e0*r)
E = ρ*a / (e0 * r)
Outside the tube, the charges of the core cancel each other.
E=0
Answer:D
Explanation:
Given
mass of object 
Distance traveled 
velocity acquired 
conserving Energy at the moment when object start falling and when it gains 12 m/s velocity
Initial Energy
Final Energy

where
is friction work if any


Since Friction is Present therefore it is a case of Open system and net external Force is zero
An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .