Answer:
a)n= 3.125 x 
electrones.
b)J= 1.515 x 
A/m²
c)
=1.114 x 
m/s
d) ver explicación
Explanation:
La corriente 'I' = 5A =>5C/s
diámetro 'd'= 2.05 x 
m
radio 'r' = d/2 => 1.025 x m
número de electrones 'n'= 8.5 x 
a) La cantidad de electrones que pasan por la bombilla cada segundo se determina mediante:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
Como sabemos que: Q= ne
donde e es la carga del electrón, es decir, 1.6 x 
C
n= Q/e => 5/ 1.6 x
n= 3.125 x electrones.
b) La densidad de corriente 'J' en el cable se calcula como
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x )²)
J= 1.515 x A/m²
c) La velocidad típica '' de un electrón se expresa como:
= 
=1.515 x / 8.5 x x |-1.6 x |
=1.114 x m/s
d) De acuerdo con estas ecuaciones,
J= I/A
= =
Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?
Answer:
consult a teacher
Explanation:
go to your school
locate the teacher
request assistance from him/her
complete the question and you're done:)
Response: Numerous elements can be found, all situated within the same vertical column as bromine.
Explanation:
Elements are organized by their atomic numbers on the periodic table. Those in the same vertical column (known as groups) exhibit the same valence electron configurations, resulting in similar chemical characteristics. Consequently, there are numerous elements sharing analogous chemical properties grouped with Bromine.
Answer:
More than 48%
Explanation:
If the interest is calculated monthly based on the outstanding balance, it leads to an effective annual rate of...
(1 + 4%)^12 - 1 = 60.1%... more than 48%
Answer:
Explanation:
In this scenario, we determine the initial velocity as follows:
The final velocity in this instance can be expressed as:
It is noted that transitioning from 7m/s to 13m/s takes 8 seconds. We can apply a specific kinematic equation to find the acceleration for the first part of the journey:
Solved for acceleration, we find:
For the subsequent route, we assume constant acceleration and that the train continues for 16 seconds, beginning with an initial velocity of 13m/s from the previous segment, allowing us to calculate the final speed via the following formula:
Substituting into the equation yields:
