calculate the magnitude of impulse applied to a 0.75 kilogram cart to change its velocity from 0.50 meters per second east to 2.
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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:
The duration, t = 3.53 seconds
Explanation:
The following information is provided:
The equation to calculate the time t is expressed as:
...... (1)
Where
s denotes the distance in feet
We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet
Substituting the value of s into equation (1) yields:
t = 3.53 seconds
Thus, the time taken by the object is 3.53 seconds, which provides the required answer.
Answer:
The excess charge is
Explanation:
According to the question, we are informed that
The diameter is
The potential of the surface is
The radius of the sphere is
by plugging in given values
Where k is Coulomb's constant with a value
Based on the question stating there are no other charges, Q represents the excess charge
Therefore
inserting the numerical values