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WARRIOR [948]
2 years ago
11

If AB has a bearing of following 234° 51' 48" and a anti-clockwise angle from AB to C is measured as 80° Calculate the bearing A

C. (Enter as numeric value of ddd.mmss e.g. 100° 20' 30" would be entered as 100.2030. Marked out of 5.00 P
Physics
1 answer:
Musya8 [376]2 years ago
4 0

Answer:

the bearing of the angle AC will be equal to 154°51' 48"

Explanation:

given,

bearing of  line AB = 234° 51' 48"

C is  anti clockwise angle from 80° from AB      

bearing of line AC = ?                  

To calculate the bearing of line AB  80° anticlockwise movement

bearing of the AC =  234° 51' 48"  - 80°                

                              = 154°51' 48"                

The bearing can be represented as 154.5148

hence, the bearing of the angle AC will be equal to 154°51' 48"

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Answer:

0.9 N

Explanation:

The electric force acting on a charge is given by:

F=qE

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q=4.5\cdot 10^{-5}C is the charge

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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

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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}})}

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Putting values in above equation, we get:

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

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

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

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