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bixtya [17]
4 years ago
11

The copper windings of a motor have a resistance of 50 Ω at 20° C when the motor is idle. After the motor has run for several ho

urs the resistance rises to 58Ω. What is the temperature of the windings now? (Ignore changes in the dimensions of the windings).
Physics
1 answer:
Drupady [299]4 years ago
5 0

Answer:

The temperature of the windings are 60.61 °C

Explanation:

Step 1: Data given

Resistance = 50 Ω

Temperature = 20.0 °C

After the motor has run for several hours the resistance rises to 58Ω.

Step 2: Calculate the new temperature

Formula: R = Rref(1 + α(T-Tref))

⇒with α = temperature coëfficiënt of Cupper at 20 °C = 0.00394/°C

⇒with Tref = reference temperature = 20°C

⇒with T = end temperature = TO BE DETERMINED

⇒with R = resistance at end temperature = 58Ω

⇒with Rref = resistance at reference temperature =  50 Ω

==> T = (R/Rref - 1)/α + Tref  

T = (58/50) - 1 )/ 0.00394 + 20

   T = 60.61 °C

The temperature of the windings are 60.61 °C

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Answer:

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I = 0.20 * 0.0784

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The wavelength of the infrared radiation is  λ = 3.174×10^{-5}m.

<h3>What is infrared radiation?</h3>

An infrared telescope is tuned to detect infrared radiation with a frequency of 9.45 THz.

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λ = 3.174 ×10^{-5} m

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1 year ago
Date<br>Page<br>What are the advantages of alcohol<br>thermometric liquid?<br>​
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Answer:

low freezing point. high vapour pressure.

<em>HOPE</em><em> </em><em>IT</em><em> </em><em>WILL</em><em> </em><em>HELP</em><em> </em><em>U</em><em>! </em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em>

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3 years ago
an object moving with uniform acceleration has a velocity of 12.ocm/s. if its x coordinate 2.00 later is 25.00cm what is its acc
Viktor [21]
<span> We're given that x=25 when t=2: </span>

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3 years ago
A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20
Klio2033 [76]

Answer:

\boxed{{\boxed{\blue{ 12.5}}}}

Explanation:

Given, for girl : Weight or force;

\rm \: F_1 = 50 \: kgf

Area of both heels;

\rm \: A_1 =  \; 2 ×1 \;  cm^2 = 2  \: cm^2

\rm \: Pressure \:  P_1  =  \cfrac{F_1}{ A_1 }  =  \dfrac{50 \: kgf}{2 \: cm {}^{2} }  = 25 \: kgf \: cm {}^{ - 1}

For elephant, Weight = Force \rm F_2 = 2000 kg•f

Area of 4 feet;

\rm \: A_2  = \; 4 \times 250 \;  cm^2 = 1000 \:  cm^2

\rm \: Pressure \:  P_2 = {F_2}/{A_2} \;  = \cfrac{2 \cancel{0 00 }\:  kgf}{1 \cancel{000} \: cm^2} =  2 \: kgf \: { \:cm}^{- 1}

Now;

\rm  = \dfrac{Pressure \:  Exerted  \: by  \: the \:  Girl}{Pressure  \: exerted  \: by \:  the  \: elephant}

=  \rm \: P_1/P_2

\implies    \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } =  \rm\cfrac{25 \:  \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

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