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telo118 [61]
4 years ago
10

If a bar magnet is falling through a loop of wire, the induced current in the loop of wire sets up a field which exerts a force

on the magnet. This force between the magnet and the loop will be attractive when:
Physics
1 answer:
Firlakuza [10]4 years ago
8 0

Answer:

When the magnet is leaving the loop

Explanation:

According to Lenz's law the direction of an induced current in a conductor will oppose the effect which produces it. As the current is induced in the wire loop and force is exerted on the magnet, the force between the magnet and the loop will be attractive when the magnet is leaving the loop because it's is the one that produces the effect which create the current.

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A wheel is rotating freely at angular speed 660 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi
sweet-ann [11.9K]

Answer:

(a)  110 rev/ min

(b) 5/6

Explanation:

As per the conservation of linear momentum,

L ( initial ) = L ( final )

I' ω' = ( I' + I'' ) ωf

I' is the rotational inertia of first wheel and I'' is the rotational inertia of second wheel which is at rest.

(a)

So,    ωf = I' ω' /  ( I' + I'' )

As I'' = 5I'

ωf = I' ω' /  ( I' + 5I' )

ωf = ω'/ 6

now we know ω' = 660 rev /  min

therefore    ωf = 660/6

                       = 110 rev/ min

(b)

Initial kinetic energy will be K'

K' = I'ω'² / 2

and final K.E. will be   K'' =  ( I' + I'' )ωf² / 2

                                   K'' = ( I' + 5I' ) (ω'/ 6)²/ 2

                                   K'' = 6I' ω'²/72

                                   K'' = I' ω'²/ 12

therefore the fraction lost is

                ΔK/K' = ( K' - K'' ) / K'

                           =  {( I'ω'² / 2) - (I' ω'²/ 12)} / ( I'ω'² / 2)

                            = 5/6

6 0
4 years ago
AN airplane travels 4000 m in 20s on a heading 0f 35 degrees north west. Calculate average velocity .
alexandr402 [8]
Isn't velocity Distance over time? if the degree isn't adding resistance it should be 4000 ÷ 20 which gives you 200mps ("per second") which is the velocity without resistance.
6 0
3 years ago
Need help with these please
Montano1993 [528]

Answer:

Explanation:

Which number all or one?

6 0
4 years ago
Read 2 more answers
How much force is required to accelerate a 5 kg mass at 20 m/s^2
Studentka2010 [4]

Hello!

\large\boxed{F = 100N}

Use the equation F = m · a (Newton's Second Law) to solve. Substitute in the given values:

F = 5 · 20

F = 100N

6 0
3 years ago
A bicycle rider has a speed of 19.0 m/s at a height of 55.0 m above sea level when he begins coasting down hill. The mass of the
lukranit [14]

Answer:

The mechanical energy of the rider at any height will be 6.34 × 10⁴ J.

Explanation:

Hi there!

The mechanical energy of the rider is calculated as the sum of the gravitational potential energy plus the kinetic energy. Since there are no dissipative forces (like friction), the mechanical energy of the rider at a height of 55.0 m above the sea level will be the same at a height of 25.0 m (or at any height), because the loss in potential energy will be compensated by a gain in kinetic energy, according to the law of conservation of energy.

Then, calculating the potential and kinetic energy at 55.0 m and 19 m/s, we can obtain the mechanical energy that will be constant:

Mechanical energy = PE + KE

Where:

PE = potential energy.

KE = kinetic energy.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the object.

g = acceleration due to gravity.

h = height.

Then, the potential energy of the rider will be:

PE = 88.0 kg · 9.81 m/s² · 55.0 m = 4.75 × 10⁴ J

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

Where "m" is the mass of the object and "v" its velocity. Then:

KE = 1/2 · 88.0 kg · (19.0 m/s)²

KE = 1.59 × 10⁴ J

The mechanical energy of the rider will be:

Mechanical energy = PE + KE = 4.75 × 10⁴ J + 1.59 × 10⁴ J = 6.34 × 10⁴ J

This mechanical energy is constant because when the rider coast down the hill, its potential energy is being converted into kinetic energy, so that the sum of potential energy plus kinetic energy remains constant.

5 0
3 years ago
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