A, Mechanical and A, Chemical.
Answer:
the block that starts moving first is block A
, fr = 1.625 N
, fr = 1.5 N
Explanation:
For this exercise we use Newton's second law, for which we take a reference system with the x axis parallel to the plane and the y axis perpendicular to the plane
X axis
fr- Wₓ = 0
fr = Wₓ
Axis y
N-
= 0
N = W_{y}
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
Cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
Wₓ = 11 sin θ
W_{y} = 11 cos θ
The equation for friction force is
fr = μ N
We substitute
μ (W cos θ) = W sin θ
μ = tan θ
We can see that the system began to move the angle.
θ = tan⁻¹ μ
So the angles are
Block A θ = tan⁻¹ 0.15
θ = 8.5º
Block B θ = tan⁻¹ 0.26
θ = 14.6º
So the block that starts moving first is block A
The friction force is
Block A
fr = Wx = W sin θ
fr = 11 sin 8.5
fr = 1.625 N
Block B
fr = 6 sin 14.6
fr = 1.5 N
Explanation:
No se me gano puntos?????????????? gracis ..m.n.b.v?????????
Answer:
Elastic potential energy store - which is the energy transferred to a spring as it's stretched.
Note that the methods applied in solving this question is the appropriate method. Check the parameters you gave in the question if you did not expect a complex number for the charges. Thanks
Answer:

Explanation:
Note: When a conducting wire was connected between the spheres, the same charge will flow through the two spheres.
The two charges were 0.65 m apart. i.e. d = 0.65 m
Force, F = 0.030 N
The force or repulsion between the two charges can be calculated using the formula:

Due to the wire connected between the two spheres, 
The sum of the charges on the two spheres = 
Note: When the conducting wire is removed, the two spheres will no longer contain similar charges but will rather share the total charge unequally
Let charge in the first sphere = 
Charge in the second sphere, q₂ = 
Force, F = 0.075 N

