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IRINA_888 [86]
3 years ago
8

If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.

Assume that the beam leaves the flashlight through a 7.0-cm aperture, that its white light has an average wavelength of 550nm, and that the beam spreads due to diffraction only. The distance from the Earth to the Moon is 384x10^3km.
Physics
1 answer:
grin007 [14]3 years ago
8 0

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

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A person weighs 60 kg. The area under the foot of the person is 150 cm2. Find the pressure exerted on the ground by the person.
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Explanation:

Applying,

P =  F/A................... Equation 1

Where P = Pressure, F = Force, A = Area.

From the question,

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F = W = mg............ Equation 2

Where m = mass of the person, g = acceleration due to gravity.

Substitute equation 2 into equation 1

P = mg/A................ Equation 3

Given: m = 60 kg, g = 10 m/s², A = 150 cm² = (150/10000) m² = 0.015 m²

Substitute these values into equation 3

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You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.
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Answer:

F=6\times 10^{-7}\ N

Explanation:

Given:

  • quantity of point charge, q=-4\times 10^{-9}\ C
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<u>We know that the electric field by the linear charge  is given as:</u>

E=\frac{\lambda}{2\pi.\epsilon_0.r}

E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}

E=150\ N.C^{-1}

<u>Now the force on the given charge can be given as:</u>

F=E.q

F=150\times 4\times 10^{-9}

F=6\times 10^{-7}\ N

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3 years ago
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