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BartSMP [9]
3 years ago
8

When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coeffic

ient of volume expansion? PLEASE HELPPP MEEEEE
Physics
1 answer:
alexandr1967 [171]3 years ago
5 0

Coefficient of volume expansion is 8.1 ×10⁻⁴ C⁻¹.

<u>Explanation:</u>

The volume expansion of a liquid is given by ΔV,

ΔV = α V₀ ΔT

ΔT = change in temperature  = 48.5° C

α =  coefficient of volume expansion =?

V₀ = initial volume = 2.35 m³

We need to find α , by plugin the given values in the equation and by rearranging the equation as,

\alpha=\frac{\Delta \mathrm{V}}{\mathrm{V}_{0} \Delta \mathrm{T}}=\frac{0.0920}{2.35 \times 48.5}=0.00081

  = 8.1 ×10⁻⁴ C⁻¹.

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1 year ago
Question C) needs to be answered, please help (physics)
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<em></em>

(b) The velocity at <em>t</em> = 2.00 s is

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<em></em>

(c) Compute the electron's position at <em>t</em> = 2.00 s:

<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>

The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:

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