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salantis [7]
3 years ago
11

Marty is developing a new way to heat an airplane's windshield to prevent it from accumulating ice when in flight. A thin layer

of clear, conductive material is injected between two layers of thick windshield plastic. A very small electric current is run through the inner material to keep the plastic layers warm. How might using a model help Marty develop his design? A. Marty could use a model to discover a new type of plastic for windshields. B. Airplane manufacturers could be sold models to use in their new planes. C. A model could be used to test his design in a lab. D. A model could be used for automobile windshields.
Physics
2 answers:
Snezhnost [94]3 years ago
8 0
The answer is : <span>A model could be used to test his design in a lab.  </span>Marty is developing a new way to heat an airplane's windshield to prevent it from accumulating ice when in flight. A thin layer of clear, conductive material is injected between two layers of thick windshield plastic. A very small electric current is run through the inner material to keep the plastic layers warm.  The model that could help Marty develop his design is the kind of model that could be used to test his design in a lab. 
stich3 [128]3 years ago
3 0

Answer:

B) A model could be used to test his design in a lab

Explanation:

The models in a laboratory scale are used usually to  test new methods or technologies (like the one Marty is developing) to see if they work or what adjustments they required to take them to real scale.

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Name the physical quantity which changes contenously during uniform-circular motion.
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3 years ago
A mixture of nitrogen and xenon gases, at a total pressure of 836 mm Hg, contains 2.80 grams of nitrogen and 24.9 grams of xenon
larisa86 [58]

Answer: Partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

p_A=p_T\times \chi_A

where,

p_A = partial pressure of substance A

p_T = total pressure

\chi_A = mole fraction of substance A

We are given:

m_{N_2}=2.80g

m_{Xe}=24.9g

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

And,

n_A=\frac{m_A}{M_A}

Mole fraction of nitrogen is given as:

\chi_{N_2}=\frac{\frac{m_{N_2}}{M_{N_2}}}{(\frac{m_{N_2}}{M_{N_2}}+\frac{m_{Xe}}{M_{Xe}})}

Molar mass of N_2 = 28 g/mol

Molar mass of Xe =  g/mol

Putting values in above equation, we get:

\chi_{N_2}=\frac{\frac{2.80}{28}}{\frac{2.80}{28}+\frac{24.9}{131}}

\chi_{N_2}=\frac{0.100}{0.100+0.190}=0.345

To calculate the mole fraction of xenon, we use the equation:

\chi_{Xe}+\chi_{N_2}=1\\\\\chi_{Xe}=1-0.345=0.655

p_{N_2}=836mmHg\times 0.345=288mmHg

p_{Xe}=836mmHg\times 0.655=548mmHg

Thus partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

6 0
4 years ago
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