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densk [106]
3 years ago
8

un conductor de plata (p=1,6x10°-6ohm x m) tiene una seccion de 5x10°-6 m°2 y una longitud de 110 m . calcular la resistencia

Physics
1 answer:
Alexus [3.1K]3 years ago
4 0

Responder:

35,2 ohm.

Explicación:

Dado:

La resistencia específica del conductor es,

La longitud del conductor es,

El área de la sección transversal del conductor es,

Sabemos que la resistencia de un conductor es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

Por lo tanto, la resistencia se puede expresar como:

R=\frac{\rho\times l}{A}

Ahora, conecte los valores dados y resuelva para 'R'. Esto da,

R=\frac{1.6\times 10^{-6}\ ohm\cdot m\times 110\ m}{5\times 10^{-6}\ m^2}\\\\R=35.2\ ohm

Por lo tanto, la resistencia del conductor es de 35,2 ohm.

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Which of the following are single-displacement reactions?
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Answer:

A & D

Explanation:

A single-displacement reaction is a chemical reaction whereby one element is substituted for another one in a compound and thereby generating a new element and also a new compound as products.

From the options, only options A & D fits this definition of single-displacement reactions.

For option D: Both left and hand and right hand sides each have one element and one compound. We can see that K is substituted from KBr to join Cl to form KCl and Br2 on the right hand side.

For option A: Both left and hand and right hand sides each have one element and one compound. We can see that OH is substituted from 2H2O to join Mg to form Mg(OH)2 and H2 on the right hand side.

The other options are not correct because they don't involve only and element and a compound on each side of the reaction.

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Constants Part A wo charged dust particles exert a force of 7.5x102 N n each other What will be the force if they are moved so t
alexandr1967 [171]

Answer:

New force, F'=48\times 10^3\ N

Explanation:

It is given that,

Force acting between two charged particles, F=7.5\times 10^2\ N

We need to find the force if they are moved so they are only one-eighth as far apart.

The force between two charged particles separated at a distance of r is given by :

F=k\dfrac{q_1q_2}{r^2}............(1)

If the charges are one-eighth as far apart then, r' =(1/8)r and new force is given by :

F'=k\dfrac{q_1q_2}{(\dfrac{r}{8})^2}..........(2)

Dividing equation (1) and (2) :

\dfrac{F}{F'}=\dfrac{1}{64}

F'=7.5\times 10^2\ N\times 64

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