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Eddi Din
1 month ago
7

You are stranded on a deserted island with 100 m of 18 gauge insulated copper wire, a light bulb rated at 3.0 V, a 50 V battery,

a large iron nail, a few simple tools (including wire cutters), and your clothes. There are of course trees, rocks, sand, and other assorted island flora and fauna. (The island is a bit unique: there is a dampening field present that inhibits fire.) Since you have been stranded on the island, each night after dark a helicopter flies over.
Using the materials at hand, describe a way to get yourself rescued. (You must support your description by including brief references to any laws or equations that apply.)
Physics
2 answers:
Keith_Richards [3.2K]1 month ago
7 0

Respuesta:Se puede obtener rescate enviando señales de SOS tanto de día como de noche como se explica a continuación

Justificación:

Para responder a la pregunta, notamos que

Durante el día

Dado que tenemos arena y rocas, podemos inscribir SOS en la arena

También podemos usar las rocas para construir tres marcas de X en un patrón triangular señalando SOS

De noche

1. Crear una luz intermitente para atraer al helicóptero

Para eso, podemos envolver el cable de cobre en una bobina, conectarlo a la bombilla de 3.0 V y a la batería de 50 V

La corriente a través de la bobina disminuye y aumenta a medida que el clavo de hierro se desliza dentro y fuera de la bobina, haciendo que la luz se atenúe y brille, de modo que intermitente.

La ecuación para la corriente a través de la bobina se da de la siguiente manera;

Current, \ I = \frac{Voltage}{Impedance}

Donde:

Impedancia = \bar Z = R +j·X

El valor de X varía a medida que el clavo se mueve en la bobina.

2. Señal SOS

La señal SOS también puede hacerse directamente encendiendo y apagando la bombilla usando tres destellos cortos seguidos de tres destellos largos y luego tres destellos cortos, que se pueden lograr completando y desconectando la bombilla eléctrica, el cable de cobre y la batería de 50 V directamente como se ha descrito o cubriendo la bombilla encendida constantemente con la ropa de forma secuencial.

Maru [3.3K]1 month ago
6 0

Explicación:

Dado que la persona tiene un cable de cobre, enciende la bombilla y una batería. Para mostrar su marca durante la noche, encendería la bombilla.

Para eso, haría dos cables. Conectar un cable con el extremo negativo de la batería y conectar el otro extremo del cable con la base de la bombilla. Conectar el otro cable al extremo positivo de la batería y a la base de la bombilla. Esto completaría el circuito para encender la bombilla.

Ahora, dado que la batería también necesita ser cargada, utilizará el cable de cobre y un clavo de hierro. Cuando mueva el clavo de hierro rápidamente a través de una bobina de cable de cobre, se producirá electricidad que se usará para cargar la batería.

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The acceleration of segment D is m/s2. Rank segments A, B , C from least accelerations to greatest acceleration. Least
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Answer:

D, C, B, A

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The derivative from a velocity-time graph provides the acceleration value.

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8 0
2 months ago
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A vector A is added to B=6i-8j. The resultant vector is in the positive x direction and has a magnitude equal to A . What is the
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The answer is letter d, 8.3.

 

Here’s a solution for the given problem:

 

We have:

B = 6i - 8j 


Let A be unknown; we'll denote A as = mi + nj 



The resultant A+B lies along the x-axis (which implies A+B = Ki + 0j, where K is yet to be determined,

 

and we also know the magnitude of A+B is equivalent to the magnitude of A,

 

therefore, mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2. 



Using vector addition, A+B becomes (m+6)i + (n-8)j.

 

Since we know A+B = Ki + 0j, we can establish that: 

m + 6 = K 


n - 8 = 0, which gives n=8. 

Thus, K^2=m^2+n^2 means (m+6)^2 = m^2 +8^2 


= m^2 + 12m + 36 = m^2 + 64 


which gives us 12m = 28 


m = 2.33333... 

Consequently, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333.

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