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?
The result is -15.625 m/s².
Acceleration signifies the alteration of velocity over a specified duration. It can be calculated with this formula:
Where:
vf = final velocity
vi = initial velocity
t = time
Let’s examine the information provided in your query:
Initially, the vehicle was traveling at 25 m/s before coming to a halt. Thus, it was in motion and subsequently ceased moving, indicating that the final velocity is 0 m/s.
However, we notice that the problem does not provide a time value. We need to determine the time taken from when it was in motion to when it reached the traffic light located 20 m away.
The time can be calculated using the kinematics equation:
We derive the equation by substituting the known values first.
The duration from when it was in motion until it stopped is 1.6s. Now we can utilize this in our acceleration calculation.
It is important to note that the acceleration is negative, indicating the vehicle slowed down.
Answer:
a) 
b) 
c) 
Explanation:
According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.
With this information, we can compute the ball’s starting speed.
a) Let's first assess the horizontal trajectory.
(1)
This gives us our initial equation.
Next, we need to examine the vertical trajectory.
Utilizing 
in our first equation (1)
Now let’s solve for t.
The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.
b) To determine the velocity magnitude just before impact, we must calculate both x and y components.
The computed velocity magnitude is:
c) The ball's angle is:
Referencing the diagram below, we can deduce from the geometry that x = 2.5 - 0.55 = 1.95 m, leading to cos θ = 1.95/2.5 = 0.78. Therefore, θ = cos⁻¹ 0.78 = 38.74°. According to the free body diagram, the tension in the chain measures 450 N. Here, F denotes the centripetal force and W signifies Dee's weight. The tension's components are as follows: Horizontal component = 450 sin(38.74°) = 281.6 N, directed to the left, and Vertical component = 450 cos(38.74°) = 351.0 N, directed upward. Answers: Horizontal: 281.6, directed left. Vertical: 351.0 N, directed upward.
