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?
Explanation:
Data provided:
Area A = 10 cm×2 cm = 20×10⁻⁴ m²
Separation distance d between the plates = 1 mm = 1×10⁻³ m
Battery voltage, or emf = 100 V
Resistance = 1025 ohm
Solution:
In an RC circuit, the voltage across the plates varies with time t. At the outset, the voltage matches that of the battery, V₀ = emf = 100V. However, after a certain time t, both the resistance and capacitance alter this, leading to a final voltage V expressed as
Applying the natural logarithm to both sides,
(1)
Next, we can determine the capacitance using the plates' area.
C = ε₀A/d
= 
= 18×10⁻¹²F
We can now find the time it takes for the voltage to drop from 100 to 55 V by substituting C, V₀, V, and R values into equation (1)
= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)
= 15×10⁻⁹s
= 15 ns
Answer:
Explanation:
Each of the processes connected to these molecules varies.
For instance, water that has accumulated in the atmosphere returns to the ground as rain. Cows utilize this water from local water sources. This represents one method in which water transitions from the atmosphere to the cow's body.
Regarding carbon and nitrogen, the air inhaled by cows contains nitrogen, oxygen, carbon dioxide, and other gases. These molecules enter the cow through respiration.
Answer:
17.35 × 10^(-6) m
Explanation:
Mass; m = 50 kg
Weight; W = 554 N
From the formula:
W = mg
This simplifies to; 554 = 50g
g = 554/50
g = 11.08 m/s²
Also, using the formula;
mg = GMm/r²
hence; g = GM/r²
Rearranging gives;
r = √(GM/g)
With G as a known constant of 6.67 × 10^(-11) Nm²/kg²
r = √(6.67 × 10^(-11) × 50/11.08)
r = 17.35 × 10^(-6) m
