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jeka94
3 years ago
13

Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length

L 2, how do L 1 and L 2 compare if wire 1 is made from copper and wire 2 is made from aluminum? The resistivity of copper is 1.7 × 10 –5 Ω·m and the resistivity of aluminum is 2.82 × 10 –5 Ω·m.
Physics
1 answer:
SVEN [57.7K]3 years ago
7 0

Answer with Explanation:

We are given that

Length of wire 1=L_1

Length of wire 2=L_2

Resistivity of copper wire=\rho_1=1.7\times 10^{-5}\Omega-m

Resistivity of aluminum wire=\rho_2=2.82\times 10^{-5}\Omega-m

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance =\frac{\rho l}{A}

When diameter of wires are same then their cross section area are also same .

l=\frac{RA}{\rho}

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity  is small then the length of wire will be large.

\rho_1

Therefore, L_1>L_2

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66

L_1=1.66L_2

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Answer:

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According to the question:

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Also, we know that the force in a magnetic field is given by:

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What is the frequency and wavelength, in nanometers, of photons capable of just ionizing nitrogen atoms?
nika2105 [10]

Answer:

The frecuency and wavelength of a photon capable to ionize the nitrogen atom are ν = 3.394×10¹⁵ s⁻¹  and λ = 88.31 nm.

Explanation:It is possible to know what are the frequency and wavelength of a photon capable to ionize the nitrogen atom using the equation of the energy of a photon described below.

E = hc/λ  (1)

Where h is the Planck constant, c is the speed of light and λ is the wavelength of the photon.

But first, it is neccesary to know the ionization energy of the nitrogen atom. The ionization energy is the energy needed to remove an electron from an atom, for the Nitrogen atom it will lose an electron of its outer orbit from the nucleus, farther snuff, so the electric force is weaker. Experimentally, it is known that it has a value of 14.04 eV. This value is easy to found in a periodic table.

So the nitrogen atom will need a photon with the energy of 14.04 eV to remove the electron from its outer orbit.

Replacing the Planck constant, the speed of light and the energy of the photon in the equation 1, the wavelength can be calculated:

λ = hc/E  (2)

Where h = 6.626×10⁻³⁴ J.s and c = 3.00×10⁸ m/s

But the Planck constant can be expressed in electron volts:

1 eV = 1.602 x 10⁻¹⁹ J

h = 6.626x10⁻³⁴ J/1.602x10⁻¹⁹ J . eV .s

h= 4.136x10⁻¹⁵ eV.s

Now, it is convenient to express the speed of light in nanometers:

1nm = 1x10⁻⁹ m

c = 3.00x10⁸ m/ 1x10⁻⁹ m

c = 3x10¹⁷ nm/s

Substituting in equation 2:

λ =  (4.136x10⁻¹⁵ eV.s)(3x10¹⁷ nm/s)/14.04 eV

λ = 1240 eV. nm/ 14.04 eV

λ = 88.31 nm

The frenquency is calculated using the equation 2 in the following way:

E = hν  (3)

Where ν is the frecuency

ν = E/h

ν = 14.04 eV/4.136×10⁻¹⁵ eV.s

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So the frecuency of a photon, capable to ionize the nitrogen atom, will be 3.394×10¹⁵ s⁻¹ and its wavelength 88.31 nm.

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The x-component of the length of the minute hand is:

d_{x}(t)=dcos(\theta (t)) (1)

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Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

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Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

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