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3 years ago

¿Cuál es el lado positivo sobre el auge de la industrialización

Physics

1 answer:

Nesterboy [21]3 years ago

3 0

Answer:

El lado positivo que tuvo el auge de la industrialización es principalmente que: Permitió el desarrollo económico de una gran cantidad de países. Marco un nuevo estilo de vida con mayor índice de globalización y producción. Creó una gran cantidad de fuentes de empleo.

Explanation:

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Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tig

pogonyaev

Answer:

The current needed is 1790.26 A

Explanation:

Given;

magnitude of magnetic field, B = 1.5 T

length of the solenoid, L = 1.8 m

diameter of the solenoid, d = 75 cm = 0.75 m

The magnetic field is given by;

$B = \frac{\mu_o NI }{L}$

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

I is current in the solenoid

N is the number of turns, calculated as;

$N = \frac{Length \ of\ solenoid}{diameter \ of \ wire} \\\\N = \frac{1.8}{1.5*10^{-3}} =1200 \ turns$

The current needed is calculated as;

$I = \frac{BL}{\mu_o N} \\\\I = \frac{1.5 *1.8}{4\pi *10^{-7} *1200} \\\\I = 1790.26 \ A$

Therefore, the current needed is 1790.26 A.

8 0

3 years ago

Read 2 more answers

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.