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balu736 [363]
3 years ago
11

What is the resistance of the coil A at 600 kelvin if its resistance at 300 kelvin is 50 ohms? (Assume the temperature coefficie

nt of resistance of the coil is .007/ degrees C)
Physics
2 answers:
leonid [27]3 years ago
8 0

155Ω

Explanation:

R = R ref ( 1 + ∝ ( T - Tref)  

where R = conduction resistance at temperature T

R ref = conductor resistance at reference temperature

∝ = temperature coefficient of resistance for conductor

T = conduction temperature in degrees Celsius

T ref = reference temperature that ∝ is specified at for the conductor material

T = 600 k - 273 k = 327 °C

Tref = 300 - 273 K = 27 °C

R = 50 Ω ( 1 + 0.007 ( 327 - 27) )

R = 155Ω

Vilka [71]3 years ago
3 0

Answer:

<em>155 ohms</em>

Explanation:

Resistance is the opposition to the flow of electric current via a conductor. The expression bellow is used in calculating the resistance of a coil given the temperature coefficient of resistance of the coil;

R = Rn [1 + α (T - Tn)]

where

R is the conductor's resistance at T temperature

Rn is the conductor's resistance at Tn temperature

α is the temperature coefficient of resistance of conducting material

T is the conductor's temperature

Tn is the conductor's reference temperature.

The conduct in question is a coil,

given  

R = ?

Rn = 50 ohms

α = 0.007/°C

T = 600 Kelvin

Tn = 300 Kelvin

Using the equation R = Rn [1 + α (T - Tn)] we substitute to find R

R = 50 [1 + 0.007 (600-300)]

using BODMAS The inner bracket comes first followed by multiplication, addition and then the outer bracket.

<em>R = 155 ohms </em>

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parallel-plate capacitor is made of two square plates 25 cm on a side and 1.0 mmapart. The capacitor is connected to a 50.0-V ba
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Answer:

6.9 x 10^-7 J  

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Explanation:

<u>Identify the unknown:  </u>

The energies stored in the capacitor before and after the plates are pulled farther apart  

<u>List the Knowns: </u>

Voltage of the battery: V = 50 V

Area of the plates: A = 0.25 x 0.25 = 0.0625 m^2

Original distance between the plates: d = 1 mm = 10^-3 m

New distance between the plates: d = 2 mm = 2 x 10^-3 m

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C=∈o*A/d

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U_c=(1/2)*V^2*C

<u>Solve the Problem:   </u>

<u>Before the plates are pulled farther apart:  </u>

C = 8.85 x 10^-12 x 0.0625/10^-3 = 5•53 x 10^-10 F  

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U_c = (1/2) x (50)^2 x 2•77 x 10^-10 = 3.5 x 10^-7 J  

The energy decrease because the capacitance decrease, so the stored charge decrease and transferred to the battery  

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