<span>The angle of reflection is C). the angle that the reflected ray makes with a line drawn perpendicular to the reflecting surface, that way the reflection can be seen.
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Answer:
The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
Explanation:
Given the data in the question;
Using the Clapeyron equation
![(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}](https://tex.z-dn.net/?f=%28%5Cfrac%7BdP%7D%7BdT%7D%20%29_%7Bsat%20%7D%20%3D%20%5Cfrac%7Bh_%7Bfg%7D%7D%7BTv_%7Bfg%7D%7D)
![(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }](https://tex.z-dn.net/?f=%28%5Cfrac%7BdP%7D%7BdT%7D%20%29_%7Bsat%20%7D%20%3D%20%5Cfrac%7B%5Cfrac%7BH_%7Bfg%7D%7D%7Bm%7D%20%7D%7BT%5Cfrac%7BV_%7Bfg%7D%7D%7Bm%7D%20%7D)
where
is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu
T is the temperature ( 15 + 460 )R
m is the mass of water ( 0.5 Ibm )
is specific volume ( 1.5 ft³ )
we substitute
/
272.98 Ibf-ft²/R
Now,
![(\frac{P_2 - P_1}{T_2 - T_1})_{sat](https://tex.z-dn.net/?f=%28%5Cfrac%7BP_2%20-%20P_1%7D%7BT_2%20-%20T_1%7D%29_%7Bsat)
where P₁ is the initial pressure ( 50 psia )
P₂ is the final pressure ( 60 psia )
T₁ is the initial temperature ( 15 + 460 )R
T₂ is the final temperature = ?
we substitute;
![= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}](https://tex.z-dn.net/?f=%3D%20%28%2015%20%2B%20460%20%29%20%2B%20%5Cfrac%7B%2860-50%29psia%28%5Cfrac%7B144in%5E2%7D%7Bft%5E2%7D%29%20%7D%7B272.98%7D)
![T_2 = 475 + 5.2751\\](https://tex.z-dn.net/?f=T_2%20%3D%20475%20%2B%205.2751%5C%5C)
480.275 R
Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
Answer:
Same frequency, shorter wavelength
Explanation:
The speed of a wave is given by
![v=f\lambda](https://tex.z-dn.net/?f=v%3Df%5Clambda)
![\lambda=\dfrac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bv%7D%7Bf%7D)
where,
f = Frequency
= Wavelength
It can be seen that the wavelength is directly proportional to the velocity.
Here the frequency of the sound does not change.
But the velocity of the sound in air is slower.
Hence, the frequency remains same and the wavelength shortens.