When a baseball player catches a baseball, the glove "gives" or moves backward. How does this movement affect the change in mome

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 derived frequency equals 2.63 Hz. Explanation: For an object weighing 8.0 kg with a spring stretching 3.6 cm, calculations involving the spring constant and oscillation frequency lead to this specific oscillation rate.

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.