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Stolb23 [73]
3 years ago
15

In the macroscopic world, you know that you can hear but cannot see around corners. Under what conditions does light bend around

corners (i.e. diffract) ? Explain why sound diffracts easily around a classroom door. 5. Suppose you added to the single slit an identical slit a distance d=0.25mm away from the first. Draw the resulting interference pattern you might expect on the same screen. What happens when we increase the distance between slits ? What happens in the limit that d becomes arbitrarily large?
Physics
1 answer:
omeli [17]3 years ago
6 0

Answer:

a much larger slit, the phenomenon of Sound diffraction that slits for light.

this is a series of equally spaced lines giving a diffraction envelope

Explanation:

The diffraction phenomenon is described by the expression

    d sin θ = m λ

Where d is the distance of the slit, m the order of diffraction that is an integer and λ the wavelength.

 

For train the diffraction phenomenon, the d / Lam ratio is decisive if this relation of the gap separation in much greater than the wavelength does not reduce the diffraction phenomenon but the phenomena of geometric optics.

The wavelength range for visible light is 4 10⁻⁷ m to 7 10⁻⁷ m. The wavelength range for sound is 17 m to 1.7 10⁻² m. Therefore, with a much larger slit, the phenomenon of Sound diffraction that slits for light.

When we add a second slit we have the diffraction of each one separated by the distance between them, when the integrals are made we arrive at the result of the interference phenomenon, a this is a series of equally spaced lines giving a diffraction envelope

When I separate the distance between the two slits a lot, the time comes when we see two individual diffraction patterns

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11.76 cm² into two significant figures and include the appropriate units
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You will need to turn this into scientific notation, so:

1.176 × 10^1 cm^2

To two significant figures means to two digits so since 7 is greater than 5, you will need to round up.

Final answer is:

1.2×10^1 cm^2
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3 years ago
On which planet would your weight be the most and the least?
hoa [83]

Answer:

Jupiter and neptune

Explanation:

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The current theory of the structure of the
Mariana [72]

Answers:

a) 2.82(10)^{21} kg

b) 1410 J

c) 36.62 m/s

Explanation:

<h3>a) Mass of the continent</h3>

Density \rho  is defined as a relation between mass m and volume V:

\rho=\frac{m}{V} (1)

Where:

\rho=2720 kg/m^{3} is the average density of the continent

m is the mass of the continent

V is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}

Finding the mass:

m=\rho V (2)

m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3}) (3)

m=2.82(10)^{21} kg (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy K is given by the following equation:

K=\frac{1}{2}mv^{2} (5)

Where:

m=2.82(10)^{21} kg is the mass of the continent

v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s is the velocity of the continent

K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2} (6)

K=1410 J (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass m=77 kg and the same kinetic energy as that of the continent 1413 J, we can find its velocity by isolating v from (5):

v=\sqrt{\frac{2 K}{m}} (6)

v=\sqrt{\frac{2 (1413 J)}{77 kg}}

Finally:

v=36.62 m/s This is the speed of the jogger

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