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Svetach [21]
3 years ago
14

The charge of an electron is -1.60x10-19 C. A current of 1 A flows in a wire carried by electrons. How many electrons pass throu

gh a cross section of the wire each second?
Physics
1 answer:
faltersainse [42]3 years ago
4 0

Answer: 6.241\times 10^{18} electrons pass through a cross section of the wire each second.

Explanation:

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

Given : Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

To calculate the charge passed we use the equation:

I=\frac{q}{t}

where,

I = current passed = 1 A

q = total charge = ?

t = time required = 1 sec

Putting values in above equation, we get:

1A=\frac{q}{1s}\\\\q=1A\times 1s=1C

When 96500C of electricity is passed , the electrons passed = 6.022\times 10^{23}

1 C of electricity is passed , the electrons passed = \frac{6.022\times 10^{23}}{96500}\times 1C=6.241\times 10^{18}

Hence, 6.241\times 10^{18} electrons pass through a cross section of the wire each second.

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nikitadnepr [17]
The moon goes in front of the sun
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5 0
3 years ago
In a home stereo system, low sound frequencies are handled by large "woofer" speakers, and high frequencies by smaller "tweeter"
Hitman42 [59]

Answer:

C = 2.9 10⁻⁵ F = 29 μF

Explanation:

In this exercise we must use that the voltage is

          V = i X

          i = V/X    

where X is the impedance of the system

in this case they ask us to treat the system as an RLC circuit in this case therefore the impedance is

          X = \sqrt{R^2 + ( wL - \frac{1}{wC})^2 }

tells us to take inductance L = 0.

The angular velocity is

         w = 2π f

the current is required to be half the current at high frequency.

Let's analyze the situation at high frequency (high angular velocity) the capacitive impedance is very small

         \frac{1}{wC} →0       when w → ∞

therefore in this frequency regime

         X₀ = \sqrt{R^2 + ( \frac{1}{2\pi  2 10^4 C} )^2 } =  R  \sqrt{ 1+ \frac{8 \ 10^{-10} }{RC}     }

the very small fraction for which we can despise it

        X₀ = R

to halve the current at f = 200 H, from equation 1 we obtain

         X = 2X₀

let's write the two equations of inductance

          X₀ = R                                    w → ∞

          X= 2X₀ = \sqrt{R^2 +( \frac{1}{wC} )^2 }        w = 2π 200

 

         

we solve the system

         2R = \sqrt{R^2 +( \frac{1}{wC} )^2 }

         4 R² = R² + 1 / (wC) ²

         1 / (wC) ² = 3 R²

          w C = \frac{1}{\sqrt{3} } \ \frac{1}{R}

          C = \frac{1}{\sqrt{3} } \ \frac{1}{wR}

           

let's calculate

           C = \frac{1}{\sqrt{3} } \ \frac{1}{2\pi  \ 200 \ 9}

           C = 2.9 10⁻⁵ F

           C = 29 μF

7 0
3 years ago
11. Velocity-time graph for the motion of an object in a straight path is a straight line parallel to the time axis
Vilka [71]

a) uniform velocity

b) zero or no acceleration

c) (see picture)

EXPLANATION:

(see picture)

8 0
2 years ago
Consider the elementary gas-phase reversible reaction A 3C Pure A enters at a temperature of 400 K and a pressure of 10 atm. At
xxMikexx [17]

Answer:

  • 39%

Explanation:

The equilibrium equation is:

          A\rightleftharpoons 3C

The initial concentration of A can be calculated from the ideal gas equation:

                 pV=nRT\\\\n/v=p/(RT)\\\\C_A=\dfrac{10atm}{0.08206(atm-dm^3/K-mol)/times 400K}\\\\C_A=0.304mol/liter

Determine the conversion of substance A to substance C using an ICE table and the Kc constant:

             A               ⇄          3C

I            0.304                        0

C             - x                          + 3x

E          0.304 - x                   3x

         K_c=0.25=\dfrac{(3x)^3}{(0.304-x)}

Solve for x:

You need to use a graphing calculator:

  • 108x³ = 0.304 - x
  • 108x³ + x - 0.304 = 0
  • x ≈ 0.1195 mol/liter

Then:

           C_A=0.304mol/liter-0.1195mol/liter=0.1845mol/liter\\\\C_C=3\times 0.304mol/liter=0.912mol/liter

The equilbrium conversion is:

           \% = [0.304mol/liter-0.1845mol/liter]/(0.304mol/liter)\times 100

           \% \approx 39\%

3 0
3 years ago
a boy pulls 23 kg box with a 100 N force at 39 above a horizontal surface if the coefficient of kinetic friciton between the box
tatuchka [14]

Answer:

Total work done is 2606.08 J.

Explanation:

Given :

Mass of box , m = 23 kg .

Force applied , F = 100 N .

Angle from horizon , \theta=39^o.

Coefficient of kinetic friction , \mu=0.16.

Distance travelled by box , d = 34 m .

Now ,

Total work done = work done by boy + work done by friction.

W=Fdcos\theta+f_sdcos\theta=Fd-\mu(mg) \\\\W=100\times cos39^o\times 34-0.16\times 23 \times 9.8 \\\\W=77.71\times 34-36.06=2606.08\ J.

Hence , this is the required solution.

6 0
3 years ago
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