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Rus_ich [418]
2 years ago
14

What would be the best design for an experiment that tests how much water expands when frozen?

Physics
1 answer:
galben [10]2 years ago
5 0

B. Purchase a small plastic container and mark 1-ounce increments on the outside to determine volume. Pour 5 ounces of water into the container, and place in the freezer for 8 hours. Compare the frozen or ending volume with the liquid or beginning volume.

<h3>How much water expands when frozen?</h3>

Ice is less denser than the liquid form. Water is the only known non-metallic substance that expands when it freezes because it is the unique property of water. Water density decreases and it expands approximately  about 9% by volume. For calculating the expansion of water, plastic container is the best option. We know that water expands when the water freezes because it is a unique property of water which allows the survival of aquatic organisms.

So we can conclude that option B is the right answer.

Learn more about water here: brainly.com/question/1313076

#SPJ1

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A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

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Answer:

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