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maw [93]
3 years ago
15

Technician A says that the evacuation process will remove dirt and debris from the refrigerant system. Technician B says that th

e evacuation process will remove moisture and air from the refrigerant system. Who is correct?
Physics
1 answer:
GuDViN [60]3 years ago
3 0

Answer: Technician B is right.

Explanation:

Evacuation process is used in refrigeration systems to remove moisture, air and non-profit condensable gases in order to achieve maximum function of the system.

vacuum pump is used to draw the sealed AC system into a vacuum. Evacuation of a refrigerant system also helps to maintain pressure, this is so as pulling a vacuum on the system is simply removing matter (mostly air and nitrogen) from inside the system so that the pressure inside drops below atmospheric pressure.

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The closely packed cones in the fovea of the eye have a diameter of about 2 μm. For the eye to discern two images on the fovea
Marina CMI [18]

Answer:

The distance between the object is l=0.0056\  cm

Explanation:

The free body diagram of this setup is on the first uploaded image

From the question

   The diameter of closely packed cones in the fovea of the eye is  =  2 \mu m

     The distance of separation by one cone(not excited ) is d = 4\mu m = 4*10^{-4}cm

     The distance between the two point-like object  is  l

     The diameter of the eye is D = 2 cm

     The distance of the two point-like object from the near point of the eye is A = 28 cm

 From the diagram we see that the light from the two point-like object form a triangle of similar base l and d  and height D and A

So for a triangle with similar base we have that

                \frac{l}{A} =\frac{d}{D}

                \frac{l}{28} = \frac{4*10^{-4}}{2}

making l the subject we have

            l = \frac{28 *4*10^{-4}}{2}

              l=0.0056\  cm

               

         

       

   

5 0
3 years ago
Read 2 more answers
(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a c
andreyandreev [35.5K]

Answer:

\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}

Explanation:

The formula for the surface of the circle is:

A(r) = \pi\cdot r^{2}

The rate of change of the spill area is obtained by deriving the previous formula in terms of time:

\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}

Finally, variables are replaced by known data:

\frac{dA}{dt} = 2\pi\cdot (39\,m)\cdot (1\,\frac{m}{s} )

\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}

4 0
4 years ago
Read 2 more answers
Which statement is correct about the car?
JulijaS [17]

Answer:

B

Explanation:

OOf we are doing this stuff atm

So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B

6 0
3 years ago
Read 2 more answers
A car engine changes chemical potential energy into the what energy so the car can move
yuradex [85]
It transforms it to mechanical
7 0
3 years ago
Two ropes support a load of 478 kg. The two ropes are perpendicular to each other, and the tension in the first rope is 2.2 time
sveta [45]

Answer:

T₂ = 1937.68 N

Explanation:

First, we will calculate the weight of the object:

W = mg = (478\ kg)(9.81\ m/s^2)\\W = 4689.18\ N

Now, we will calculate the resultant tension in the ropes. Since the ropes are perpendicular. Therefore,

T = \sqrt{T_1^2+T_2^2}\\

where,

T = Resultant Tension

T₁ = Tension in rope 1

T₂ = Tension in rope 2

According to the given condition tension in the first rope is 2.2 times the tension in the second rope:

T₁ = 2.2 T₂

Therefore

T = \sqrt{(2.2T_2)^2 + T_2^2}\\\\T =  2.42T_2

Now, the weight of the object must be equal to the resultant tension for equilibrium:

T = W\\2.42T_2 = 4689.18\ N\\\\

<u>T₂ = 1937.68 N</u>

5 0
3 years ago
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