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dusya [7]
3 years ago
7

If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?

Physics
1 answer:
Klio2033 [76]3 years ago
5 0
From tables, the density of mercury is
13545 kg/m^3 at 20°C,
13472 kg/m^3 at 50°C.

Because mass = density * volume, the mass of mercury at 20°C is
m = (13545 kg/m^3)*(0.002 m^3) = 27.09 kg

Let V  = volume of mercury at 50°C.
Because the mass of mercury does not change, therefore at 50°C,
(13472 kg/m^3)*(V m^3) = 27.09
V = 27.09/13472 = 0.0020108 m^3

Answer:  B. 0.002010812 m³
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Please solve the Problem.
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  • The equivalent capacitance between point a and b is 5.87μf.
  • The charge at 20μf is 93.92 μC.
  • The charge at 6μf is 67.8 μC.
  • The charge at C and 3μf is 26.12 μC.

<h3>Sum of capacitance of C and 3μF</h3>

The sum of the capacitance is calculated as follows;

1/Ct = 1/C + 1/3

1/Ct = 1/10μf + 1/3μf

1/Ct = (3μf + 10μf)/30μf²

1/Ct = 13μf/30μf²

Ct = 30μf²/13μf

Ct = 2.31μf

<h3>Total capacitance in parallel arrangement</h3>

The total capacitance in parallel arrangement is calculated as follows;

Ct = 2.31μf + 6μf = 8.31μf

<h3>Equivalent capacitance between point a and b</h3>

1/Ct = 1/8.31μf + 1/20μf

1/Ct = 0.1703

Ct = 5.87μf

<h3>Charge flowing in each capacitor</h3>

Maximum voltage is delivered in 20μf, q = CV

<u>charge for 20μf:</u>

q = (5.87 x 16)μC

q = 93.92 μC

<h3>Equivalent capacitance for C, 3μf and 6μf</h3>

Ct = 2.31μf + 6μf = 8.31uf

<u>charge for 6μf:</u>

q = (6/8.31) x 93.92μC

q = 67.8μC

<h3>Total charge for C and 3μf</h3>

q = 93.92μC - 67.8μC = 26.12 μC

charge for C = charge 3μf = 26.12 μC

Learn more about capacitor here: brainly.com/question/14883923

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