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Ratling [72]
1 year ago
12

what current flows through a 2.54cm diameter rod of pure silicon that is 20.0cm long ,when 1.00x10000V is applied to it

Physics
1 answer:
Andrej [43]1 year ago
8 0

0.04 Ampere of current will flow through the silicon.

<h3>What is Electrical Current ?</h3>

Electrical current is the rate at which charges flow in the conductor. It is measured in Ampere (A)

What current flows through a 2.54cm diameter rod of pure silicon that is 20.0cm long ,when 1.00x10000V is applied to it

The given diameter is 2.54 cm, the radius r = 2.54/2 = 1.27 cm

Convert cm to m

  • r = 1.27/100 = 0.0127m
  • L = 20/100 = 0.2m

Area A = πr²

A = π × 0.0127²

A = 0.000507 m²

The resistivity ρ of silicon material is 640 Ωm

Resistance R =  ρL / A

R = ( 640 × 0.2 ) / 0.000506

R = 128 / 0.000506

R = 252964.43 Ω

From Ohm's law

V = IR

10000 = 252964.43I

I = 10000 / 252964.43

I = 0.0395A

Therefore, 0.04 Ampere of current will flow through the silicon.

Learn more about Current here: brainly.com/question/24858512

#SPJ1

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6 0
2 years ago
An organ pipe of length 3.0 m has one end closed. The longest and next-longest possible wavelengths for standing waves inside th
iris [78.8K]

Answer:

The longest wavelength for closed at one end and open at the other is

y / 4      where y is the wavelength - that is node - antinode

The next possible wavelength is 3 y / 4 -    node - antinode - node -antinode

y / 4 = 3 m     y = 12 meters    the longest wavelength

3 y / 4 = 3 m      y = 4 meters   1 / 3 times as long

6 0
3 years ago
A 295-kg object and a 595-kg object are separated by 4.10 m.
kodGreya [7K]

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

F_{13}=\dfrac{Gm_1m_3}{r^2}

by putting the values

F_{13}=\dfrac{Gm_1m_3}{r^2}

F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}

F₁₃=2.94 x 10⁻⁷ N

The force  between the mass m₂ and m₃

by putting the values

F_{23}=\dfrac{Gm_2m_3}{r^2}

F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

F_{23}=\dfrac{Gm_2m_3}{x^2}

F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}

Form above two equation

\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}

\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}

\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

4 0
2 years ago
Please help! I’ve got 96 N but it’s wrong..
mash [69]

Dude it’s 8.1 Km. Bet it’s right.

5 0
3 years ago
Read 2 more answers
In the diagrams, the solid green line represents Earth's rotational axis and the dashed red line represents the magnetic field a
Elis [28]

Answer:

b

Explanation:

5 0
3 years ago
Read 2 more answers
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