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SIZIF [17.4K]
2 years ago
6

A cryogenic vacuum pump works by condensing vapors onto some absorbent medium. This is an efficient and clean way to pump a syst

em in a research environment. The term cryo means cold, which indicates that these types of vacuum pumps contain a refrigerant cycle to cool the internal parts. The temperature difference between the inside and outside of a typical cryogenic pump is Δ=303 ∘C . Derive an expression to convert this difference into Fahrenheit and express the answer.
Physics
1 answer:
Julli [10]2 years ago
6 0

Answer:  Temperature in Fahrenheit is 577.4

Explanation:

The conversion factor for converting celcius to Fahrenheit is:

F=\frac{9}{5}\times C+32

where F = temperature in Fahrenheit

C = Temperature in Celcius

Given : Temperature difference in Celcius = 303^0C

Putting in the values we get:

F=\frac{9}{5}\times 303+32

F=577.4

Thus the answer in Fahrenheit is 577.4

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Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the
Hoochie [10]

You've told us:

-- 130°x  =  212°F

and

-- 10°x  =  32°F

Thank you.  Those are two points on a graph of °x vs °F .  With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are  (130, 212)  and  (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation.  I'll use the second point  (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are !  Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad.  Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for  x, and the solution will be the same temperature in  F  too.

or

We can write [ F = 1.5F + 17 ], solve it for  F, and the solution will be the same temperature in  x  too.

F = 1.5F + 17

Subtract  F  from each side:  0.5F + 17 = 0

Subtract 17 from each side:   0.5F = -17

Multiply each side by 2 :  F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for  x:  

F = 1.5(-34) + 17

F = -51 + 17

<em>F = -34</em>

It works !  -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

<em>yay !</em>

6 0
3 years ago
What step does the Rin PRICES stand for? Why is this step important?
posledela

Explanation:

The five-step process for treating a muscle or joint injury such as an ankle sprain is called "P.R.I.C.E." which is short for Protection, Rest, Ice, Compression, and Elevation).

6 0
3 years ago
What is the total distance that the object traveled?
Natasha_Volkova [10]

Answer:

for what?

Explanation:

d=S x T

or

d=vt+1/2at2

srry if wrong but

hope this helps

take care

8 0
3 years ago
A circular plate of 500-mm diameter is maintained at T1 = 600 K and is positioned coaxial to a conical shape. The back side of t
drek231 [11]
For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K

For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W
5 0
3 years ago
Consider the following situations: i. A ball moving at speed v is brought to rest ii. The same ball starts at rest and is projec
tatyana61 [14]

Answer:

c. Case iii

Explanation:

the ball will experience the largest change in case iii

3 0
3 years ago
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