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Sav [38]
3 years ago
10

A 20 centimeter long copper wire sensor is designed to alert if the temperature elevates beyond a safe region causing the wire t

o make contact with the wall, 1.5 mm away at 25 C. What temperature (in K) will the copper wire make contact with the wall, setting the sensor off? The thermal coefficient of expansion, α, for copper (Cu) is found to be 16.6 ppm/K (16.6x10-6 /K).
Physics
1 answer:
Dennis_Churaev [7]3 years ago
4 0

Answer:

The temperature is 749.8 K

Explanation:

Final temperature (T2) = (distance apart/thermal coefficient of expansion×length) + initial temperature

distance apart = 1.5 mm = 1.5/1000 = 0.0015 m

thermal coefficient of expansion for copper = 16.6×10^-6/K

Length of copper = 20 cm = 20/100 = 0.2 m

Initial temperature = 25 °C = 25 + 273 = 298 K

T2 = (0.0015/16.6×10^-6×0.2) + 298 = 451.8 + 298 = 749.8 K

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What is the relative size and composition of the universe, a galaxy, and a solar system?
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2 years ago
Which type of electronic exchange connects buyers with sellers within a specific industry?
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5 0
3 years ago
Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

4 0
4 years ago
Help with question 16 and 17
storchak [24]
<h3>16.</h3>

Your answer is correct.

___

<h3>17.</h3>

The fractional change in resistance is equal to the given temperature coefficient multiplied by the change in temperature.

  R = R₀×(1 + α×ΔT)

  R = (10.0 Ω)×(1 + 0.004×(65 -20)) = 11.8 Ω

5 0
4 years ago
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