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torisob [31]
3 years ago
7

A steel pot has a bottom with a cross sectional area of .1m2 and has a thickness of 1cm. The pot is filled with boiling water an

d the bottom of the pot is held at 200 degrees Celsius. The thermal conductivity of steel is 14 J/s m Co. How much heat is being conducted through the pot in one minute?
Physics
1 answer:
jenyasd209 [6]3 years ago
3 0

Answer:

840000 J/min

Explanation:

Area = A = 0.1 m²

Bottom of pot temperature = 200 °C

Thermal conductivity = k = 14 J/sm°C

Thickness = L = 1 cm = 0.01 m

Temperature of boiling water = 100 °C

From the law of heat conduction

Q = kAΔT/L

⇒ Q = 14×0.1×(200-100)/0.01

⇒ Q = 14000 J/s

Converting to J/minute

Q = 14000×60 = 840000 J/min

∴ Heat being conducted through the pot is 840000 J/min

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The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5
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Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

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where K = 8.99e9

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Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) =  7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) =  4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.89e5)^{2}  + (7.89e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) net magnitude =  1.115e6\frac{N}{C} @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

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E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.88e5)^{2}  + (7.88e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) and (b) the net magnitude =  1.242e6\frac{N}{C} @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

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