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dmitriy555 [2]
10 months ago
7

The human body has an average density of 979 kg/m3 , what fraction of a person is submerged when floating gently in fresh water?

Physics
1 answer:
FromTheMoon [43]10 months ago
4 0

A person is submerged of about 97.9%.

The average density of the human body is given as 979 kg/ m³.

<h3>Define Law of floatation.</h3>

    Law of floatation can be defined as the volume of the liquid displaced when a body floats on the liquid surface is equal to the body submerged in the water.

As body has the stable equilibrium state, the buoyancy of the fluid will be equal to the weight.

Weight of the body floating = Weight of the body immersed in fluid

  Law of floatation = Density of the floating object / density of fluid

 As fluid is the freshwater here, the density of fluid will be 1000 kg/ m³.

                               = (979 kg/ m³) / ( 1000 kg/ m³)

                               = 97.9 %

A person is submerged when floating gently in fresh water about 97.9%.

Learn more about law of floatation,

brainly.com/question/17032479

#SPJ4

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N
Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

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A spherical weather balloon is filled with hydrogen until its radius is 4.40 m. Its total mass including the instruments it carr
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Answer:

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4329.10484 N

Explanation:

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The Buoyant force = Weight of the air displaced

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