Answer:
Acceleration = 2.35 m/
Speed = 8.67 m/s
Explanation:
The coefficient of friction , u =0.3
The angle of incline = 30°
The two forces acting on block are weight and friction.
weight along the incline = mg cos60° = = 0.5 mg
Friction along incline = umg cos30° = mg
Friction along incline = 0.26 mg
Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg
Acceleration = = 0.24 g = 2.35 m/
The height of incline = 8 m
Length of the inclined edge = 16 m
v= 8.67 m/s
Complete Question
The complete question is shown on the uploaded image
Answer:
The nuclei experience a force that will move it to the right of the conductor rod while the electrons experience a force that will move it to the left side of the conductor rod.
Explanation:
The force that act on the charges(both the positive and the negative charge ) is Mathematically expressed as
F = qE => E = F/q
where F is the force, q is the charge and E is the electric field
we can see that the force is directly proportional to the electric field,so an increase in the electric field would increase the force.
we can also see that the electric field is given as force per unit charge and generally the direction of this field is taken to be the direction of the force it would exert on a positive test charge
Now from the question we are being told that the external electric field is the direction of the positive x axis
Hence this field would drive the positive charge i.e the nuclei to the right.
In order to further explain let consider this
Generally the electric field is always radially outward originating from a positive point charge and radially in toward a negative point charge.
what this means for this question, is that the positive point charge is on the left side of the electric field while the negative point charge is at the right side of the field.
According to Coulomb's law which states that unlike term attract while like terms repel, it means that the electron would move to the left of the conductor rod while the nuclei would move to the right side of the conductor rod.
