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UkoKoshka [18]
3 years ago
14

What is an item in which magnetic domains can be aligned and a magnetic field induced for a short period of time

Physics
2 answers:
dusya [7]3 years ago
6 0

Answer:temporary magnets

(soft magnets) retain their magnetic properties for a short time

example: paper clips and nails become magnetized by being near or rubbing against a magnet

Leto [7]3 years ago
3 0
I think an object such as an iron pipe could work
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I’m steel, the solvent is And the solute is. .
sleet_krkn [62]
Iron is the solvent and carbon is the solute.
8 0
3 years ago
A rotating flywheel has moment of inertia 18.0 kg⋅m^2 for an axis along the axle about which the wheel is rotating. Initially th
timama [110]

Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

Explanation:

By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:

\Delta E = K_{1}-K_{2} (1)

Where:

\Delta E - Energy losses, measured in joules.

K_{1}, K_{2} - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2}) (2)

Where:

I - Moment of inertia of the flywheel, measured in kilograms per square meter.

\omega_{1}, \omega_{2} - Initial and final angular speed, measured in radians per second.

If we know that K_{1} = 30\,J, K_{2} = 15\,J and I = 18\,kg\cdot m^{2}, then the initial angular speed is:

K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2} (3)

\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }

\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }

\omega_{1} \approx 1.825\,\frac{rad}{s}

\omega_{1}\approx 0.291\,\frac{rev}{s}

K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2} (4)

\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }

\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }

\omega_{2} \approx 1.291\,\frac{rad}{s}

\omega_{2} \approx 0.205\,\frac{rev}{s}

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

t = \frac{\omega_{2}-\omega_{1}}{\alpha} (5)

If we know that \omega_{1}\approx 0.291\,\frac{rev}{s}, \omega_{2} \approx 0.205\,\frac{rev}{s} and \alpha = -0.200\,\frac{rev}{s^{2}}, then the time needed is:

t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }

t = 0.43\,s

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

6 0
3 years ago
How does a mathematical model help you understand the science concepts? Give an example.
sveta [45]
Usually describes a system by a set of variables in a set of equations established relationships between the variables and variables maybe of many types real or integer numbers Boolean values of strings for example
4 0
3 years ago
The weight of a person can be represented by a vector that acts...
Maurinko [17]
Straight down, even if the person is standing on a hill
4 0
3 years ago
A 600 kg car is at rest, and then accelerates to 5 m/s.
Crank

Answer:

0

7500J

7500J

Explanation:

Given parameters:

Mass of car  = 600kg

Velocity  = 5m/s

Unknown:

Original kinetic energy  = ?

Final kinetic energy  = ?

Work used  = ?

Solution:

The kinetic energy of a body is the energy due to the motion of a body.

It can be solve mathematically using expression below;

                K.E  = \frac{1}{2}  m v²

  where m is mass

              v is velocity

original kinetic energy;

  The car started at rest and v = 0, therefore K.E  = 0

Final kinetic energy;

           K.E  = \frac{1}{2}  x 600 x 5²   = 7500J

Work done:

   Work done  = Final K.E  - Initial K.E  = 7500 - 0 = 7500J

   

3 0
3 years ago
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