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2 years ago

Uranus (mass = 8.68 x 1025 kg) and

Physics

1 answer:

Pepsi [2]2 years ago

5 0

Answer:129,398,203.7 m

Explanation:

According to Newton's law of Universal Gravitation, the force $F$ exerted between two bodies of masses $M$ and $m$ and separated by a distance $d$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{Mm}{d^2}$ (1)

Where:

$F=2.28(10)^{19} N$ is the gravitational force

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$M=8.68(10)^{25} kg$ is the mass of Uranus

$m=6.59(10)^{19} kg$ is the mass of Uranu's moon, Mirana

$d$ is the distance between Uranus and its moon

Isolating $d$ :

$d=\sqrt{\frac{GMm}{F}}$ (2)

$d=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.68(10)^{25} kg)(6.59(10)^{19} kg)}{2.28(10)^{19} N}}$ (3)

Finally:

$d=129,398,203.7 m$

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During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

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Answer:

part a : The dry unit weight is 0.0616  $lb/in^3$ 

part b : The void ratio is 0.77

part c : Degree of Saturation is 0.43

part d : Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

The dry unit weight is 0.0616  $lb/in^3$ 

Part b

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The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

The void ratio is 0.77

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

Degree of Saturation is 0.43

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

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w is the moisture content in percentage, given as 12% or 0.12 in fraction

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Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

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