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3 years ago

8

Por un alambre de cobre circula una intensidad de corriente de 5 A en 120 segundos, ¿Cuál es la cantidad de carga que circula po

r el alambre de cobre

1 answer:

wolverine [178]3 years ago

3 0

Answer:

?>

Explanation:

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Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of

sweet [91]

Answer:

$1.44\times 10^{-3}$ N

Explanation:

We are given that three charged particle are placed at each corner of equilateral triangle.

$q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC$

$q_1=-8.2\times 10^{-9} C$

$q_2=-16.4\times 10^{-9} C$

$q_3=8.0\times 10^{-9} C$

Side of equilateral triangle =3.3 cm= $\frac{3.3}{100}=0.033m$

We know that each angle of equilateral angle= $60^{\circ}$

Net force=F = $\sum\frac{kQq }{d^2}$

Where k= $9\times 10^9 Nm^2/C^2$

If we bisect the angle at $q_3$ then we have 30 degrees from there to either charge.

Direction of vertical force due to charge $q_1$ and $q_2$

Therefore, force will be added

Vertical force= $9\times 10^9\times q_3(q_1+q_2)\frac{cos30}{(0.033)^2})$

Vertical net force= $9\times 10^9\times 8\times 10^{-9}(-8.2-16.4)\times 10^{-9}\times 10^6\times\frac{\sqrt3}{2\times 1089}$

Vertical force = $9\times 8(-24.6)\times 10^{-9}\times 10^6\times 1.732\times \frac{1}{2178}$

Vertical force= $-1.41\times 10^{-3}N$ (towards $q_1}$

Horizontal component are opposite in direction then will b subtracted

Horizontal force= $9\times 10^9\times 8\times 10^{-9}(-8.2+16.4) \times 10^{-9}\times \frac{sin30}{(0.033)^2}$

Horizontal force= $0.27\times 10^-3}$ N(towards $q_2$

Net electric force acting on particle 3 due to particle = $\sqrt{F^2_x+F^2_y}$

Net force= $\sqrt{(-1.41\times 10^{-3})^2+(0.27\times 10^{-3})^2}$

Net force= $1.44\times 10^{-3}$ N

3 0

3 years ago

A force of 50 n is exerted to the right on a 15 kg car at the same time a 30 n force is exerted to the left on the car. find the

choli [55]

Find out what the force is and yiy have your answer

5 0

4 years ago

there is only one basic format for a professional report. a. audience b. All of these should be considered when deciding on a re

Anna007 [38]

Answer:

The answer is letter b. All of these should be considered when deciding on a report format.

Explanation:

A Professional Report is a type of formal document about a topic or information that is intended for a specific audience or purpose. The report's style of writing needs a lot of knowledge from the writer. Oftentimes, it involves the following important elements: <em>Title, Summary, Body, Discussion, Conclusion and Recommendation. </em>

The writer should write according to his target audience and purpose. He also needs to consider the length of his report, as well as the suitable words and sentences that he should use.

Thus, all of the choices are important in writing a professional report. So, the answer is letter b.

5 0

3 years ago

A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

8 0

4 years ago

a cheerleader throws herself into the air with a velocity of 12 m/s at an angle of 75 degrees above the horizontal. what are the

NISA [10]

Answer:

Vx = 3.10 [m/s]

Vy = 11.59 [m/s]

Explanation:

To solve this problem we must decompose the velocity vector by means of the angle on the horizontal.

v = 12 [m/s]

Vx = 12*cos (75) = 3.10 [m/s]

Vy = 12*sin (75) = 11.59 [m/s]

7 0

4 years ago