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adoni [48]
3 years ago
8

Suppose you have designed a new thermometer called the x thermometer. on the x scale the boiling point of water is 129 ∘x and th

e freezing point of water is 12 ∘x. part a at what temperature are the readings on the fahrenheit and x thermometers the same? express your answer using two significant figures.
Physics
1 answer:
gladu [14]3 years ago
8 0

For any temperature scale the ratio of difference of temperature will be same

so we can write

\frac{T^ox - 12^ox}{129^ox - 12^ox} = \frac{T^of - 32^of}{212^of - 32^of}

now given that in both scales the temperature must be same

\frac{T^ox - 12^ox}{117^ox} = \frac{T^of - 32^of}{180^of}

T - 12 = 0.65 (T - 32)

0.35T = -8.8

T = -25.14 ^ox

<em>so at above temperature both scales will have same temperature</em>

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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
jenyasd209 [6]

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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3 years ago
The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separ
alexdok [17]

Answer:

U₁ = (ϵAV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

Explanation:

The energy stored in a capacitor is given by (1/2) (CV²)

Energy in the capacitor initially

U = CV²/2

V = voltage across the plates of the capacitor

C = capacitance of the capacitor

But the capacitance of a capacitor depends on the geometry of the capacitor is given by

C = ϵA/d

ϵ = Absolute permissivity of the dielectric material

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U = ϵAV²/2d

Now, for U₁, the new distance between plates, d₁ = 3d

U₁ = ϵAV²/2d₁

U₁ = ϵAV²/(2(3d))

U₁ = (ϵAV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

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