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statuscvo [17]
3 years ago
11

Two points along a wire are labeled Xand Y. The current is measured to be iXY= –3A.The reference direction of iXY is defined by

double-subscript notation. a. In which direction are electrons flowing? From X to Y, or from Y to X? b. How much charge (in coulombs) passes through a cross section of the wire in 5 seconds?
Engineering
1 answer:
klasskru [66]3 years ago
4 0

Answer:

a. From Y to X

b. q = 15 C

Explanation:

a.

When current is denoted by double subscript, it is interpreted as traveling from 1st subscript to the second, when the value of current is positive. On the other hand, if the value of current is negative, then it means that the current is traveling from 2nd subscript to the 1st subscript. Since, the value of current is negative in the given question, therefore, it means that the current is traveling from 2nd subscript to the 1st subscript. Hence the direction of current or the flow of electrons is:

<u>From Y to X</u>

<u></u>

b.

Using the following formula of current:

I = q/t

where,

I = Current (Absolute Value) = 3  A

q = amount of charge = ?

t = time taken = 5 s

Therefore,

3 A = q/5 s

q = (3 A)(5 s)

<u>q = 15 C</u>

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El tiempo hasta que falle un sistema informático sigue una distribución Exponencial con media de 600hs. (Utilice 3 decimales par
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Answer:

La probabilidad pedida es 0.717

Explanation:

Primero comencemos definiendo la variable aleatoria. Para nuestro problema, la variable aleatoria es la siguiente :

X: '' El tiempo (en horas) hasta que falle un sistema informático ''

La variable aleatoria X será entonces una variable aleatoria continua.

Sabemos que sigue una distribución exponencial con una media de 600 hs.

Esto se escribe :

X ~ ε ( λ ) (I)

En donde λ es igual a la inversa de la media. Esto se escribe :

λ =\frac{1}{E(X)}

En donde E(X) es la media de la variable. Por ende, si reemplazamos los datos del ejercicio obtenemos ⇒

λ =\frac{1}{E(X)} ⇒ λ =\frac{1}{600}

Si reemplazamos el valor de λ en (I) obtenemos :

X ~ ε (\frac{1}{600})

La función de distribución de X (por ser una variable aleatoria exponencial) es :

F_{X}(x)=P(X\leq x)=  1 - e ^ ( - λx) con x > 0 y F_{X}(x)=0 en caso contrario.

Si reemplazamos el valor de λ en la función de distribución de X obtenemos :

F_{X}(x)=P(X\leq x)=1-e^{-\frac{x}{600}}  

Dado que la variable aleatoria X se distribuye de manera exponencial, el hecho de saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos aporta ninguna información sobre lo que ocurrirá después. Esta característica se conoce como propiedad de perdida de memoria de la variable aleatoria exponencial. Entonces, la probabilidad pedida se reduce a calcular :

P(X>200)    

Dado que saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos dice nada sobre lo que ocurrirá instantes posteriores a esas 400 hs.

Calculamos entonces la probabilidad pedida :

P(X>200)=1-P(X\leq 200)=1-F_{X}(200)=1-(1-e^{-\frac{200}{600}})=e^{-\frac{1}{3}}=0.717

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Answer:

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