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3 years ago

What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface

Physics

1 answer:

kipiarov [429]3 years ago

4 0

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman $= \frac{95.08}{100 } = 0.9508$

The density of the woman is calculate as;

$Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water$

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

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4 years ago

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[Refraction]

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Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

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3 years ago

Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water. If the cylinder

jonny [76]

Answer:

0.10839 m

Explanation:

$P_1$ = Atmospheric pressure = 1 atm = 101325 Pa

$P$ = Total pressure at bottom of mecury = 1.2 atm

g = Acceleration due to gravity = 9.81 m/s²

h = d = Depth of mercury

$\rho_m$ = Density of mercury = $1.36\times 10^4\ kg/m^3$

$\rho_w$ = Density of water = $1000\ kg/m^3$

Pressure at the bottom is of the cylinder is given by

$P_2=P_1+\rho_wgh\\\Rightarrow P_2=101325+1000\times 9.81(0.7-d)$

Pressure at the bottom of mercury is

$P=P_2+\rho_mgh\\\Rightarrow 1.2\times 101325=101325+1000\times 9.81(0.7-d)+1.36\times 10^4\times 9.81\times d\\\Rightarrow 1.2\times 101325=123606d+108192\\\Rightarrow d=\dfrac{1.2\times 101325-108192}{123606}\\\Rightarrow d=0.10839\ m$

The depth of the mercury is 0.10839 m

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3 years ago

John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2

Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function $cos(\omega t +\phi)$ is defined between -1 and 1, So it is not possible obtain a value $2\pi$ greater.

In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have $f=1+cos(\omega t +\phi)$ , the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the $f=1+cos(\omega t +\phi)$ , and its mean that the function $f=cos(\omega t +\phi)$ , in general, is not greater than $cos(\omega t +\phi)$ .

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3 years ago