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

FromTheMoon [43]

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,

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4 0

1 year ago

A wire of resistance 5.9 Ω is connected to a battery whose emf ε is 4.0 V and whose internal resistance is 1.2 Ω. In 2.9 min, ho

dexar [7]

Answer:

a) 390J

b) 322J

c) 68J

Explanation:

We need to calculate the power given by the battery. the power is given by:

$P=V*I\\I=\frac{V}{R}\\I=\frac{4}{5.9+1.2}\\I=0.56A\\P=2.24W$

Watts is J/s so:

$E=P*t\\E=2.24\frac{J}{s}*2.9min*(60\frac{s}{min})=390J$

The thermal energy in the wire is given by:

$E_w=P_w*t\\P_w=I^2*R_w=0.56^2*5.9=1.85W\\E_w=1.85\frac{J}{s}*2.9min*(60\frac{s}{min})=322J$

And the the dissipated thermal energy in the battery will be the remainig energy:

$E_b=E-E_w\\E_b=390-322=68J$

4 0

3 years ago

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

$I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}$

$I =mr^2(0.287)$

I = 0.287 MR²

3 0

3 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c

EleoNora [17]

Answer:

63.8 V

Explanation:

We are given that

$A_1=1.6 cm^2=1.6\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

$A_2=1.2 cm^2=1.2\times 10^{-4} m^2$

$A_3=4.4 cm^2=4.4\times 10^{-4} m^2$

$A_4=7 cm^2=7\times 10^{-4} m^2$

Potential difference,V=140 V

We know that

$R=\frac{\rho l}{A}$

According to question

$l_1=l_2=l_3=l_4=l$

In series

$R=R_1+R_2+R_3+R_4$

$R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})$

$R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})$

$R=\rho l(18284.6)$

$I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}$

Potential across 1.2 square cm= $V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V$

Hence, the voltage across the 1.2 square cm wire=63.8 V

3 0

3 years ago

Fertilization occurs in the fallopian tubes.

Tamiku [17]

What’s the question? Is it true or false?

7 0

3 years ago