Let’s first be less vague. We’ll say pressure, there’s an intensive property. Also, let’s identify a material. We’ll say water as its vapor behavior is so well understood.
So, if what you are asking is would the pressure of saturated water vapor that boiled off of pure water be different than the pressure of saturated water vapor that was generated from a solution (let’s not require it be saturated) of, say, water and sugar that was mostly sugar, if the temperatures are equal then the answer is yes. Why? Because the boiling points are different.
For example, if you were to find saturated vapor in a system at atmospheric pressure above water, then that steam would be at 212 °F and 0 psig (14.7 psia). Water vapor above a 50%+ DS (dissolved solid to water by mass percentage) solution of dextrose and water would have a partial pressure lower than atmospheric in the same setting. Why? Because we still have a ways to go before that solution will actually boil.
Now, when you do reach its boiling point you will have a vapor that is at 0 psig but it will not be saturated. It will instead be superheated, as the dextrose molecules have a negligible vapor pressure. The vapor you get is basically all water (with some tiny entrained but not vaporized sugar droplets), and as the boiling point was greater than the saturated temperature for that pressure it’s superheated out of the gate. If you were to take it out of that system and allow it to desuperheat on its own (without the addition of a water spray) it would lose pressure to achieve saturation.
I hope that helped.
Answer:
The answer is 0.727
Explanation:
lemme know if that's right
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Explain : I've got a lot of help from this website :D
Answer:
10.984mm
Explanation:
by elastic modulus
stress=modulus of elasticity*strain
stress=loading/area area" cross-section"
11mm=0.011m
area=π(d/2)^2=π(0.011/2)^2=9.503*10^-5 square meter
stress=55000/(9.503*10^-5)=578.745 MPa
convert MPa and GPa to pascal.
strain=stress/modulus=(578.745*10^6)/(125*10^9)=0.00463............axial strain
v=Poisson ratio
lateral strain=(-v)*axial strain= -0.31*0.00463
lateral strain= -1.4353*10^-3=change in diameter/ original diameter
change in diameter=(-1.4353*10^-3)*0.011= -1.57883*10^-5 m
negative indicates decrease in diameter.
decrease in dia.=0.01578mm
new diameter=11-0.01578= 10.984mm
Answer: It will be saved $81 per year.
Explanation: A <u>car without the tires</u>, that gets 35mpg and is driven 15,000 miles per year, will have a fuel cost of:
1 gal ---- 35 miles
x gal ---- 15,000 miles
x =
x = 428.57 gal
Per year, 1 gal costs $3.5, then:
cost = 428.57*3.5
cost = 1500
<u>The car with the tires</u>, improve the efficiency in nearly 2mpg, i.e.:
efficiency = 35 + 2 = 37mpg
1 gal ----- 37 miles
y gal ----- 15,000 miles
y =
y = 405.40 gal
Per year:
cost = 405.40*3.5
cost = 1419
The difference in costs is:
d = 1500 - 1419
d = 81
With lowest rolling resistance tires, a car will save, per year, $81.