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Elza [17]
3 years ago
15

What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m?

Physics
1 answer:
polet [3.4K]3 years ago
4 0

Answer:

0.044 V

Explanation:

E = Electric field = 5.5\times 10^6\ V/m

d = Thickness of membrane = 8 nm

When the electric field strength is multiplied by the membrane thickness we get the voltage

Voltage across a gap is given by

V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V

The voltage across the membrane is 0.044 V

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A gas is compressed at a constant pressure of 0.800 atm from 12.00 L to 3.00 L. In the process, 390 J of energy leaves the gas b
Andru [333]

Answer:

a)W= - 720 J

b)ΔU= 330 J

Explanation:

Given that

P = 0.8 atm

We know that 1 atm = 100 KPa

P = 80 KPa

V₁ = 12 L = 0.012 m³       ( 1000 L = 1 m³)

V₂ = 3 L = 0.003 m³

Q= - 390 J ( heat is leaving from the system )

We know that work done by gas given as

W = P (V₂ -V₁ )

W= 80 x ( 0.003 - 0.012 ) KJ

W= - 0.72 KJ

W= - 720 J    ( Negative sign indicates work done on the gas)

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in the internal energy

Now by putting the values

- 390 = - 720 + ΔU

ΔU= 720 - 390  J

ΔU= 330 J

5 0
3 years ago
The average Intensity of the electromagnetic field due to sunlight at the surface of the earth is 1400 W/m^2. Calculate the maxi
Ann [662]

Answer:

1027 N/C

3.42 x 10⁻⁶ T

Explanation:

I = Intensity of electromagnetic field = 1400 W/m²

E₀ = Maximum value of electric field

Intensity of electromagnetic field is given as

I = (0.5) ε₀ E₀² c

1400 =  (0.5) (8.85 x 10⁻¹²) (3 x 10⁸) E₀²

E₀ = 1027 N/C

B₀ = maximum value of magnetic field

using the equation

E₀ = B₀ c

1027 = B₀ (3 x 10⁸)

B₀ = 3.42 x 10⁻⁶ T

4 0
3 years ago
The Moon is attracted to the Earth. The mass of the Earth is 6.0x1024 kg and the mass of the Moon is 7.4x1022 kg. If the Earth a
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Sorry I don't know the answer

7 0
3 years ago
Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

l^2 = 2.8286m^2

l =\sqrt{2.8286m^2}

l = 1.68m

Therefore the length of the wire is 1.68m

6 0
3 years ago
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D. <span>Johannes Kepler argued that Earth was the center of the universe.

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4 0
4 years ago
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