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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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PLEASE PLEASE I DONT KNOW HOW TO ANSWER THIS ANYTHING WOULD HELP THANKS SO MUCH
fomenos

Thankfull for point's.

8 0
3 years ago
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You're conducting a physics experiment on another planet. You drop a rock from a height of 2.1 m and it hits the ground 0.6 seco
umka21 [38]

The formula to use is the one that connects the acceleration,
the distance fallen, and the time spent falling:

         Distance = 1/2 a T² .

You said  2.1 meters in 0.6 second .

                                                         2.1 m              = 1/2 a  (0.6 sec)²

Multiply each side by  2 :                  4.2 m              =       a (0.6 sec)²

Divide each side by (0.6 sec)² =     (4.2/0.36) m/s² =        a

                                                         a = (11 and 2/3) m/s²

                                        (about 19% more than Earth's gravity)

3 0
3 years ago
Read 2 more answers
A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t
Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

(C) at 33.23 m the speed will be 12 m/s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height  is given by;

v = u - gt

u = gt

t = \frac{u}{g}

Time of flight, T = 2t

T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s

(C) the position of the ball at 12 m/s

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0
3 years ago
What is caused by the condensation of water vapor?
Lisa [10]
Condensation<span> is the change of </span>water<span> from its gaseous form (</span>water vapor) into liquid water<span>. </span>Condensation<span> occurs in the atmosphere when warm air rises, cools and looses its capacity to hold </span>water vapor<span>. As a result, excess </span>water vapor <span>condenses to form cloud droplets.</span>
7 0
3 years ago
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An electron travels through a particular point in an experimental apparatus with a velocity of 0.949 \times 10^6×10 ​6 ​​ m/s an
Rzqust [24]

Answer:

The magnitude of the acceleration of the electron at this point is 3.94\times10^{14}\ m/s^2.

Explanation:

Given that,  

Velocity v= 0.949\times10^{6}\ m/s

Angle = 61.5°

Magnetic field = 0.01 T

We need to calculate the magnetic force

Using formula of magnetic force

F=Bev\cos\theta

Where, B = magnetic field

v = velocity

e = charge of electron

Put the value into the formula

F=0.01\times1.6\times10^{-19}\times0.949\times10^{6}\cos61.5

F=3.59\times10^{-16}\ N

We need to calculate the acceleration

Using newton's second law

F= ma

a = \dfrac{F}{m}

Put the value into the formula

a=\dfrac{3.59\times10^{-16}}{9.1\times10^{-31}}

a=3.94\times10^{14}\ m/s^2

Hence, The magnitude of the acceleration of the electron at this point is 3.94\times10^{14}\ m/s^2.

5 0
3 years ago
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