Answer:
1 m
Explanation:
L = 100 m
A = 1 mm^2 = 1 x 10^-6 m^2
Y = 1 x 10^11 N/m^2
F = 1000 N
Let the cable stretch be ΔL.
By the formula of Young's modulus
ΔL = 1 m
Thus, the cable stretches by 1 m.
Four springs are stretched to the same distance from the equilibrium position. The spring constants are listed in the table. A 2
1 answer:
You might be interested in
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.
Answer:
a.
b.
c.
Explanation:
First, look at the picture to understand the problem before to solve it.
a. d1 = 1.1 mm
Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:
To solve the equations we have to convert all units to those of the international system. (mm→m)
μ0 is the constant of proportionality
μ0=4πX10^-7 N*s2/c^2
b. d2=3.6 mm
Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:
J: current density
c: outer radius
b: inner radius
The cilinder's current is negative, as it goes on opposite direction than the wire's current.
c. d3=7.4 mm
Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:
As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.
