3 years ago

According to the cooling curve, what is the approximate freezing point of the substance?

Physics

1 answer:

Oksana_A [137]3 years ago

3 0

Answer:

I do not know which substance you are referring to, but the freezing point of water is 32°F, or 0°C.

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The potential difference comes out to be

$10 \times 10 {}^{ - 3} m$

Given:

σ = 8. 85 × 10-9 c/m2

we know,

$E = \frac{σ}{2ε0}$

$E = \frac{8.85 \times 10 {}^{ - 9} }{2ε0}$

$E = \frac{v}{d}$

given the potential difference between two equipotential surface=5v

E=∆v

∆d=∆v/E

$= \frac{5 \times 8.85 \times 10 { }^{ - 12} \times 2 }{8.85 \times 10 {}^{ - 9} }$

$Δ = 10 \times 10 {}^{ - 3} m$

Thus the potential difference is

$10 \times 10 {}^{ - 3} m$

Learn more about potential difference from here: brainly.com/question/28165869

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