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2 years ago

A photon has a wavelength of 500nm what is the energy associated

1 answer:

8 0

Given: Plank's constant h = 6.626 x 10⁻³⁴ J.s

Speed of light c = 2.998 x 10⁸ m/s

Wavelength λ = 500 nm convert to meter = 5 x 10⁻⁷ m

Required: Energy E =?

Formula: E = hc/λ

E = (6.626 X 10⁻³⁴ J.s)(2.998 X 10⁸ m/s)/5 x 10⁻⁷ m

E = 3.97 X 10⁻¹⁹ J or in termS of electron volt

E = 2.48 eV

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2 years ago

Arrange the following in order of their appearance in the life cycle of a star: white dwarf, red giant, main-sequence star. Expl

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Main-sequence star, red giant, and white dwarf

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2 years ago

The drawing shows four sheets of polarizing material, each with its transmission axis oriented differently. Light that is polari

Brrunno [24]

Answer:

Explanation:

When sheet A is removed

$I_B=32\cos^230=24W/m^2\\\\I_C=24 \cos^260=6W/m^2\\\\I_D=6\cos^230=4.5W/m^2$

When sheet B is removed

$I_A=32\cos^20=32W/m^2\\\\I_C=24 \cos^290=0W/m^2\\\\I_D=0\cos^230=0W/m^2$

When sheet C is removed

$I_A=32\cos^20=32W/m^2\\\\I_D=32 \cos^230=24W/m^2\\\\I_B=24\cos^290=0W/m^2$

When sheet D is removed

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8 0

3 years ago

A hollow non-conducting spherical shell has inner radius R1 = 5 cm and outer radius R2 = 19 cm. A charge Q = -35 nC lies at the

Gnom [1K]

Answer:

a. E = -13.8 kN/C

b. E = +8.51 kN/C

Explanation:

We will apply Gauss' Law to the regions where the electric field is asked.

Gauss' Law states that if you draw an imaginary surface enclosing a charge distribution, then the electric field through the imaginary surface is equal to the total charge enclosed by this surface divided by electric permittivity.

$\int\vec{E}d\vec{a} = \frac{Q_{\rm enc}}{\epsilon_0}$

a. For this case, we will draw the imaginary surface between the inner and outer shell of the sphere. The total charge enclosed by this surface will be equal to the sum of the charge Q at the center and charge of the shell within the volume from R1 and r.

Here, r = 0.5(R1+R2) = 12 cm.

$E4\pi r^2 = \frac{Q_{\rm enc}}{\epsilon_0}\\Q_{\rm enc} = Q + \rho V_{\rm enc} = Q + (Ar) (\frac{4}{3}\pi (r^3 - R_1^3)) = (-35\times 10^{-9}) + (16\times 10^{-6})(12\times 10^{-2})(\frac{4}{3}\pi((12\times 10^{-2})^3 - (5\times 10^{-2})^3)) = -2.21\times 10^{-8}~C$

$E = \frac{-2.21\times 10^{-8}}{4\pi (12\times10^{-2})^2 \epsilon_0} = -1.38\times 10^4~N/C\\E = -13.8~kN/C$

b. For this case, we will draw the imaginary surface on the outside of the shell.

The total charge enclosed by this surface will be equal to the sum of the charge at the center and the total charge of the shell.

$Q_{\rm enc} = Q + \rho V = Q + (Ar)[\frac{4}{3}\pi (R_2^3 - R_1^3)]\\Q_{\rm enc} = (-35\times 10^{-9}) + [(16\times 10^{-6})(38\times 10^{-2})][\frac{4}{3}\pi((19\times 10^{-2})^3 - (5\times 10^{-2})^3)]\\Q_{\rm enc} = 1.36\times 10^{-7}~C$

$E = \frac{1.36\times 10^{-7}}{4\pi (38\times10^{-2})^2 \epsilon_0} = 8.51\times 10^3~N/C\\E = 8.51~kN/C$

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3 years ago

3. When one plate descends, (“subducts”, or to put it simply goes beneath another plate) parts of it are recycled into the plate

jolli1 [7]

Answer:

subduction zone

Explanation:

The answer is actually in the question itself..."subducts", or to put it simple goes beneath another plate".

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2 years ago