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2 years ago

Does passing a magnet through a coil of wire break off it’s electric current

1 answer:

7 0

A magnetic field is actually generated by a moving current (or moving electric charge specifically). The magnetic field generated by a moving current can be found by using the right hand rule, point your right thumb in the direction of current flow, then the wrap of your fingers will tell you what direction the magnetic field is. In the case of current traveling up a wire, the magnetic field generated will encircle the wire. Similarly electromagnets work by having a wire coil, and causing current to spin in a circle, generating a magnetic field perpendicular to the current flow (again right hand rule).

So if you were to take a permenant magnet and cut a hole in it then string a straight wire through it... my guess is nothing too interesting would happen. The two different magnetic fields might ineteract in a peculiar way, but nothing too fascinating, perhaps if you give me more context as to what you might think would happen or what made you come up with this question I could help.

Source: Bachelor's degree in Physics.

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A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0

Kazeer [188]

Answer:

$\Delta K = 52J$

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: $\Delta K=K_f-K_i$

We know that the formula for the kinetic energy for an object is:

$K=\frac{mv^2}{2}$

where <em>m </em>is the mass of the object and <em>v</em> its velocity.

For our case then we have:

$\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}$

Which for our values is:

$\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J$

3 0

3 years ago

An object travels in a circular path of radius 5.0 meters at a uniform speed of 10. m/s. What is the magnitude of the object's a

vladimir1956 [14]

I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s

8 0

3 years ago

Read 2 more answers

How is Coulomb’s law similar to newton’s law of gravitational force? How is it different

natulia [17]

The similarities and the differences between gravitational and electric force are listed below

Explanation:

- The magnitude of the gravitational force between two objects is given by Newton's law of gravitation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

- Coloumb's law gives instead the strength of the electrostatic force between two charged objects, which is

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

By comparing the two equations, we find the following similarities:

Both the forces are inversely proportional to the square of the distance between the two objects, $F\propto \frac{1}{r^2}$
Both the forces are proportional to the product between the "main quantity" of each force, which is the mass for the gravitational force ( $F\propto m_1 m_2$ ) and the charge for the electric force ( $F\propto q_1 q_2$

Instead, we have the following differences:

The gravitational force is always attractive, since the sign of $m$ is always positive, while the electric force can be either attractive or repulsive, since the sign of $q$ can be either positive or negative
The value of the gravitational costant G is much smaller than the value of the Coulomb's constant, so the gravitational force is much weaker than the electric force