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klasskru [66]
3 years ago
5

A large spool in an electrician's workshop has 65 m of insulation-coated wire coiled around it. When the electrician connects a

battery to the ends of the spooled wire, the resulting current is 1.8 A. Some weeks later, after cutting off various lengths of wire for use in repairs, the electrician finds that the spooled wire carries a 2.9-A current when the same battery is connected to it. What is the length of wire remaining on the spool
Physics
1 answer:
Art [367]3 years ago
8 0

Answer:

40.34\ \text{m}

Explanation:

L_1 = Length of wire = 65 m

I_1 = Initial current = 1.8 A

I_2 = Final current = 2.9 A

We know

R\propto \dfrac{1}{I}

and

R\propto L

\dfrac{V}{I}\propto L\\\Rightarrow L\propto \dfrac{1}{I}

so

\dfrac{L_2}{L_1}=\dfrac{I_1}{I_2}\\\Rightarrow L_2=\dfrac{I_1}{I_2}L_1\\\Rightarrow L_2=\dfrac{1.8}{2.9}\times 65\\\Rightarrow L_2=40.34\ \text{m}

The length of the wire remaining on the spool is 40.34\ \text{m}.

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Two people push on the same door from opposite sides as shown.
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Se apunta un rifle horizontalmente con mira a un blanco pequeño que está a 200m en el suelo. La velocidad inicial de la bala es
vfiekz [6]

Answer:

Lo importante a tener en cuenta sobre esta pregunta es que la velocidad horizontal de la bala no hace ninguna diferencia en cuanto al tiempo que tarda en caer al suelo.

Debido a que el arma no ha aplicado ninguna fuerza vertical a la bala, la única fuerza que afecta la bala es la gravedad. Esto significa que la bala tarda tanto en caer al suelo como lo haría si se cayera, a pesar de que ahora viaja una gran distancia horizontal en la duración.

Para encontrar el tiempo de viaje antes de tocar el suelo, tenemos 3 valores:

-El desplazamiento desde el suelo que la bala debe viajar, s = 1.5m

-La aceleración que experimenta la bala. Como la gravedad está acelerando la bala hacia abajo, a = g = ~ 9.81m / s ^ 2

-La velocidad inicial de la bala verticalmente. Como la bala es estacionaria verticalmente (solo viaja horizontalmente al inicio), u = 0m

Examinamos nuestras ecuaciones de movimiento, comúnmente conocidas como ecuaciones SUVAT. Es posible que necesite aprender estos para su examen, pero algunas tablas de examen los proporcionan.

Debido a que tenemos s, u y a, y estamos buscando el tiempo t, la ecuación relevante es

s = ut + 0.5 (en ^ 2)

Completando nuestros valores tenemos:

1.5 = 0t + 0.5 (9.81 x t ^ 2)

1.5 = 4.905 x t ^ 2

Divide 1.5 entre 4.905 para encontrar t ^ 2

t ^ 2 = 0.3058 ...

Simplemente encontramos la raíz cuadrada de t ^ 2 para encontrar t, el tiempo que tarda la bala en llegar al suelo:

t = 0.553s (3 cifras significativas)

Para encontrar la distancia horizontal, d, que la bala ha viajado antes de tocar el suelo, podemos usar la ecuación que vincula el desplazamiento s con cierta velocidad v durante un tiempo t:

s = vt

La velocidad horizontal de la bala, v = 430

El tiempo antes de que la bala toque el suelo, t = 0.553

Entonces d = vt = 430 * 0.553 = 238m (3 cifras significativas)

3 0
3 years ago
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