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Nonamiya [84]
3 years ago
12

A. How does increasing the voltage of a power supply in an electromagnet affect the strength of the magnetic field?

Physics
2 answers:
ahrayia [7]3 years ago
8 0

a) The electromagnetic field will become stronger. More current - more field!

b) Mechanical energy helps to move the generator's shaft. Then electromagnetic energy creates a magnetic field that crosses generator's windings with high frequency. According to Faraday's law change in magnetic flux causes voltage.

c) On the secondary side it increases voltage, decreases current, power remains the same relatively to the primary side.

d) F=qvB=0.0006*2.8*10^5*4.21=707.28 N. The force will look down.

seropon [69]3 years ago
7 0

PART A)

If we increase the voltage supply in an electromagnet then it will increase the current that is flowing in it

So here due to increase in current there will be increase in the magnetic field due to that electromagnet

PART B)

Here in electric generator the current is produced by rotating a coil between two strong magnets.

So here mechanical energy of rotation of coil is converted into electromagnetic energy.

PART C)

Step up transformer convert the lower voltage input into higher voltage output

here number of turns of coil in output side or secondary number of coils is more than the number of coils in primary side or input side

PART D)

Force on a moving charge is given by

F = q(\vec v \times \vec B)

here we know that

q = 0.000600 C

v = 2.8 \times 10^5 m/s

B = 4.21 T

now from above equation we have

F = 0.0006(2.8\times 10^5)(4.21)

F = 707.3 N

direction of force is given by right hand thumb rule

using that rule we got force downwards

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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.
MA_775_DIABLO [31]

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

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The answer is A. Hope this helps you with your work.
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