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Sholpan [36]
2 years ago
12

How can you make the potential energy as high as possible in a magnetic field between one electromagnet and one piece of iron?

Physics
1 answer:
harkovskaia [24]2 years ago
7 0

In step 1, to increase the potential energy, the iron will move towards the electromagnet.

In step 2, to increase the potential energy, the iron will move towards the electromagnet.

<h3>Potential energy of a system of magnetic dipole</h3>

The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

U = \mu B

where;

  • \mu is the dipole moment
  • B is the magnetic field

B = \frac{\mu_0 I}{2\pi r}

U = \mu\times  (\frac{\mu_0 I}{2\pi r} )

Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;

  • In step 1, to increase the potential energy, the iron will move towards the electromagnet.
  • In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.

Learn more about potential energy in magnetic field here: brainly.com/question/14383738

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A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a
katen-ka-za [31]

Answer:b

Explanation:

Given

mass of first cart m_1=6 kg

mass of second cart m_2=3 kg

velocity of first cart v_1=3 m/s

conserving momentum

m_1v_1+m_2v_2=(m_1+m_2)v

6\times 3+3\times 0=(9)\cdot v

v=2 m/s

Initial kinetic Energy K.E._1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2

K.E._1=\frac{1}{2}\cdot 6\cdot 3^2+0

K.E._1=27 J

Final Kinetic Energy

K.E._2=\frac{1}{2}(m_1+m_2)v^2

K.E._2=\frac{1}{2}(6+3)\cdot 2^2=18 J

Ratio of initial Kinetic Energy to the Final Kinetic Energy

=\frac{27}{18}=1.5

6 0
3 years ago
Why do we use the name electromagnetism to describe the force of an
ElenaW [278]

Answer:

magnet, gravity

Explanation:

5 0
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A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k
Arte-miy333 [17]

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

I_1=m_1r^2

I_1=7\times (0.25)^2=0.437\ kgm^2

The moment of inertia of the rod about one end is given by :

I_2=\dfrac{m_2l^2}{3}

l = r

I_2=\dfrac{m_2r^2}{3}

I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2

For 6 spokes, I_2=0.025\times 6=0.15\ kgm^2

So, the net moment of inertia of the wagon is :

I=I_1+I_2

I=0.437+0.15=0.587\ kgm^2

So, the moment of inertia of the wagon wheel for rotation about its axis is 0.587\ kgm^2. Hence, this is the required solution.

4 0
3 years ago
A shaving or makeup mirror is designed to magnify your face by a factor of 1.40 when your face is placed 20.0cm in front of it
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Answer:

(a) convex mirror

(b) virtual and magnified

(c) 23.3 cm

Explanation:

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distance of object, u = - 20 cm

magnification, m = 1.4

(a) As the image is magnified and virtual , so the mirror is convex in nature.

(b) The image is virtual and magnified.

(c) Let the distance of image is v.

Use the formula of magnification.

m =-\frac{v}{u}\\1.4=-\frac{v}{-20}\\v =28 cm

Use the mirror equation, let the focal length is f.

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Radius of curvature, R = 2 f = 2 x 11.67 = 23.3 cm

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