Conclusion. What process is responsible for the bubbling action of the organism? What is your evidence?

Engineering

1 answer:

noname [10]3 years ago

6 0

Answer:

Explanation:

Hands-on Activity Bubbling Plants Experiment to Quantify Photosynthesis ... After running the experiment, students pool their data to get a large sample ... Explain that photosynthesis is a process that plants use to convert light ... Describe a simple experiment that provides indirect evidence that photosynthesis is occurring.

Through photosynthesis, certain organisms convert solar energy (sunlight) into ... of our planet continuously and is transferred from one organism to another. Therefore, directly or indirectly, the process of photosynthesis provides most of the energy ... Biology in Action ... Chlorophyll is responsible for the green color of plants.Photosynthetic organisms capture energy from the sun and matter from the air to ... oxygen produced during photosynthesis makes leaf bits float like bubbles in water. ... their ability to carry out photosynthesis, the biochemical process of capturing ... this air is forced out and replaced with solution, causing the leaves to sink.

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3 years ago

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A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u

Darina [25.2K]

Answer:

The pressure upstream and downstream of a shock wave are related as

$\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}$

where,

$\gamma$ = Specific Heat ratio of air

M = Mach number upstream

We know that $\gamma _{air}=1.4$

Applying values we get

$\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)$

Similarly the temperature downstream is obtained by the relation

$\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}$

Applying values we get

$\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C$

The Mach number downstream is obtained by the relation

$M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616$