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Yuri [45]
3 years ago
15

Technician A says a refrigerant identifier is only needed when the technician is unsure of the refrigerant used in an AC system.

Technician B says some refrigerant identifiers can measure the amount of moisture in the refrigerant system to aid in diagnosis. Who is​ correct?
Physics
1 answer:
Yanka [14]3 years ago
8 0

Question:

The options are;

A) Technician A only B) Technician B only

C) Both technicians A and B D) Neither technician A nor B

Answer:

The correct option is;

D) Neither technician A nor B

Explanation:

To avoid costly errors, it is important to make use of a refrigerant identifier both on a vehicle A/C system requires work and at the market before buying a refrigerant.

There are refrigerant identifiers that can detect air contamination the however a refrigerant identifier is meant to identify the refrigerant present in the system based on purity and quality and can give an indication of the type of contamination present. All contaminated refrigerants are disposed of.

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A _____ is a quantity that has magnitude and direction
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vector is the answer of this blank

7 0
3 years ago
Which of theses substances would be a poor conductor of electricity
lys-0071 [83]

Answer:

B. Water and sugar.

Explanation:

In the given options water and sugar would be the poor conductor of electricity. Other given options such as water and salt, water and Hcl and water and NaOH are better conductor of electricity because Hcl ,NaOH, salt (Nacl) can break into their ionic form whereas water and sugar will not.

8 0
3 years ago
A solenoid 50-cm long with a radius of 5.0 cm has 800 turns. You find that it carries a current of 10 A. The magnetic flux throu
ioda

Answer:

126 mWb

Explanation:

Given that:

length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.

We assume that the magnetic field in the solenoid is constant.

The magnetic flux is given as:

\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb

8 0
3 years ago
An aluminum cup contains 225 g of water and a 40 g copper stirrer, all at 27°C. A 410 g sample of silver at an initial temperatu
fomenos

Answer:

130.22 g

Explanation:

Parameters given:

Mass of water Mw = 225 g

Mass of stirrer Ms = 40 g

Mass of silver M(S) = 410 g

By applying the law of conservation of energy:

(McCc + MsCs + MwCw)ΔTw = M(S)C(S)ΔT(S)

where Mc = Mass of cup

Cc = Specific heat capacity of aluminium cup = 900 J/gC

Cs = Specific heat capacity of copper stirrer = 387 J/gC

Cw = Specific heat capacity of water = 4186 J/gC

ΔTw = change in temperature of water = 32 - 27 = 5 °C

C(S) = Specific heat capacity of silver = 234 J/gC

ΔT(S) = change in temperature of silver = 88 - 32 = 56 °C

Therefore:

[(Mc * 900) + (40 * 387) + (225 * 4186)] * 5 = 410 * 234 * 56

(900Mc + 957330) * 5 = 5276700

900Mc + 957330 = 5276700 / 5 = 1074528

900Mc = 1074528 - 957330

900Mc = 117198

Mc = 117198/ 900

Mc = 130.22 g

The mass of the cup is 130.22 g.

5 0
3 years ago
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