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o-na [289]
3 years ago
10

A wire 1 mm in diameter is connected to one end of a wire of the same material 2 mm in diameter of twice the length. A voltage s

ource is connected to the wires and a current is passed through the wires. If it takes time T for the average conduction electron to traverse the 1-mm wire, how long does it take for such an electron to traverse the 2-mm wire
Physics
1 answer:
miskamm [114]3 years ago
8 0

Answer:

T = 2 T₀

Explanation:

To answer this question let's write the expression for electrical conductivity

    σ = n e2 τ / m*

The relationship with resistivity is

       ρ = 1 /σ

Whereby the resistance

        R = ρ L / A = 1 /σ  L / A

We see that there is no explicit relationship between time and resistance, there is only a dependence on the life time (τ) that depends on the properties of the material, not on its diameter or length.

As also the average velocity or electron velocity of electrons is constant, the time to cross 2 mm in length is twice as long as the time to cross a mm in length

 T = 2 T₀

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A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?
rjkz [21]

Answer:

Maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2              

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec

Maximum acceleration is given by a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2

So maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2

8 0
3 years ago
Read 2 more answers
En un rio una Onda viaja con una velocidad de propagación de 50 m/s con una longitud de Onda de 40 metros. Hallar la frecuencia
sattari [20]

Answer:

Frequencia = 1.25 Hz

Explanation:

<u>Dados los siguientes datos;</u>

  • Velocidad = 50 m/s
  • Longitud de onda = 40 metros

Para encontrar la frecuencia de la onda;

Matemáticamente, la velocidad de una onda viene dada por la fórmula;

Velocidad = Longitud \; de \; onda * Frequencia

Haciendo de la frecuencia el tema de la fórmula, tenemos;

Frequencia = \frac {Velocidad}{Longitud \; de \; onda}

Sustituyendo en la fórmula, tenemos;

Frequencia = \frac {50}{40}

<em>Frequencia = 1.25 Hz</em>

7 0
3 years ago
Calculate the pressure on a man’s foot when a woman who weighs 520 N steps on his foot with her heel which has an area of 0.001
Vinvika [58]

Answer:

520000  or 520000 pa

Force = 520N

Area of contact = 0.001

Pressure: 520000 or 520000

6 0
3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

6 0
3 years ago
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A light spring of constant 176 N/m rests vertically on the bottom of a large beaker of water.
irakobra [83]

Answer:

14.0 cm

Explanation:

Draw a free body diagram of the block.  There are three forces: weight force mg pulling down, elastic force k∆L pulling down, and buoyancy ρVg pushing up.

Sum of forces in the y direction:

∑F = ma

ρVg − mg − k∆L = 0

(1000 kg/m³) (4.63 kg / 648 kg/m³) (9.8 m/s²) − (4.63 kg) (9.8 m/s²) − (176 N/m) ∆L = 0

∆L = 0.140 m

∆L = 14.0 cm

8 0
3 years ago
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