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3 years ago

To protect a material from the influence of an external magnetic field, the material should be kept in?

1 answer:

5 0

To protect a material from the influence of an external magnetic field, the material should be kept in soft iron ring.

So the correct answer is A.

Hope this helps,

Davinia.

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You live on a planet far from ours. "Based on extensive communication with a physicist on earth", you have determined that all l

Ugo [173]

Answer:

8.56 m/s2

Explanation:

Using law of energy conservation while taking into account of the rotational and translation kinetic energy, when the solid cylinder rolls down the incline we have the potential energy converted to kinetic energy:

$E_p = E_{kv} + E_{k\omega}$

$mgh = mv^2/2 + I\omega^2/2$

where m is the mass, $I = mr^2/2$ is the moments of inertia of the solid cylinder $\omega = v / r$ is the angular speed of the cylinder

$mgh = mv^2/2 + \frac{mr^2}{2}\frac{(v/r)^2}{2}$

$mgh = mv^2/2 + mv^2/4 = 3mv^2/4$

$h = 3v^2/(4g)$

So if you plot a liner chart of h vs $v^2$ and get a slope of 6.42 then that means 3/(4g) = 6.42 so $g = 6.42*4/3 = 8.56 m/s^2$

The gravitational acceleration on this planet is 8.56 m/s2

3 0

3 years ago

La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.

KATRIN_1 [288]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

3 0

3 years ago

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If there was an sudden change in exchange rates that resulted in fewer U.S. Dollars being required to buy a Euro, you could expe

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4 years ago

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