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amm1812
3 years ago
8

Questions

Physics
1 answer:
Butoxors [25]3 years ago
8 0

Answer:

1. The tightness of the wire has no effect on the strength of the electromagnet

2. The strength increases with the number of coils

3. The strength of the electromagnet increases with the number of dry cells used

4. The strength of the electromagnet increases with the wideness of the nail but not the length of the nail

Explanation:

The strength of an electromagnet is given by the following relation;

B = \dfrac{\mu_0 \cdot K \cdot N \cdot I}{L}

Where;

B = The magnetic field strength at the center

μ₀ = The magnetic permeability of free space = 4·π × 10⁻⁷ N·A⁻²

N = The number of loops formed by the conductor around the core

I = The current flowing through wire coiled around the nail

K = The magnetic permeability of the nail

L = The length of the coil

Therefore, we have;

1. From the above equation, the tightness of the wire coil around the nail (or the radius, 'R', of the wire) does not does not affect the magnetic field strength

2. The number of coils, 'N', is directly related to the magnetic field strength, 'B', and therefore, increasing the number of turns or coils around the nail, increases the magnetic field strength

3. The current in the circuit is directly related to the magnetic field strength and the number of dry cell used increases the current in the circuit and therefore, can increase the magnetic field strength

4. The size of the nail used in a solenoid and the magnetic field strength are directly related. The wider the nail, the stronger the magnetic field

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An air-track glider attached to a spring oscillates between the 13.0 cm mark and the 63.0 cm mark on the track. The glider compl
LuckyWell [14K]

Answer:

(A) = 2.13 s

(B) = 0.47 Hz

(C) = 0.25 m

(D) = 0.74 m/s

Explanation:

number of oscillations (n) = 15

time (t) = 32 secs

start point (L1) = 13 cm = 0.13 m

End point (L2) = 63 cm = 0.63 m

(A) period = time / number of oscillation

           = 32 / 15 = 2.13 s

(B) frequency = 1 / period

              = 1 / 2.13 = 0.47 Hz

(C) Amplitude = 0.5 ( L2 - L1 )

                           = 0.5 ( 0.63 - 0.13 )

                          = 0.25 m

(D) max speed = (2π / T) x A

                        =  (2π / 2.13) x 0.25

                       = 0.74 m/s

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3 years ago
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A plant was placed in the corner of a room away from a window. What might you observe about the plant after five days?
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Answer:

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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.
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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