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Contact [7]
3 years ago
6

How is the surface area of an average person is 2m^2 in the chapter pressure

Physics
2 answers:
iren [92.7K]3 years ago
6 0

Answer:

The total atmospheric pressure experienced by the person's whole body is, P = 202650 N/m²

Explanation:

Given data,

The surface area of the person, A = 2 m²

The standard atmospheric pressure is  P = 101325 N/m² (or), Pascal

This is the amount of pressure experienced by 1 m² body at sea level.

If the surface area of the person is, A = 2 m², then the pressure becomes,

                                  P = 2 x 101325 N/m²

                                    = 202650 N/m²

Hence the total atmospheric pressure experienced by the person's whole body is, P = 202650 N/m²

11Alexandr11 [23.1K]3 years ago
3 0

Answer:

We have to show surface area =2m^2,with few conditions that is by considering Force =200000\ N and Pressure =100000\ Pa to be respectively.

Explanation:

The atmospheric pressure is =10^{5}\ Pa on Earth's surface.

The magnitude of the force exerted on a person by the atmosphere is =2\times 10^{5}\N (or)\ 200kN\ (or)\ 20\ ton.

Now to calculate surface area we can find it from pressure=\frac{force}{area} and re-arranging it to.

area=\frac{force}{pressure}

So plugging the values,

Surface area =\frac{20000}{10000}=2\ m^{2}

Hence from the above calculations we can say that surface area is 2m^2.

So the surface area of an average person can be said to have 2m^2, using the concept of pressure and force.

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Answer:

317.22

Explanation:

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= 0.4217

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Total

L = 109.4+ 7.93+ 36.04+ 163.85

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\boxed{ \bold{ \huge{ \boxed{ \sf{see \: below}}}}}

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\underline{ \bold{ \sf{To \: prove \: that \: kinetic \: energy =  \frac{1}{2} m {v}^{2} }}}

Let us consider, a body of mass ' m ' is lying at rest ( initial velocity = 0 ) on a smooth surface. Let a constant force F displaces this body in its own direction by a displacement ' d '. Let 'v' be it's final velocity. The work done ' W ' by the force is given by :

\sf{W = FD}

⇒\sf{W = m \:  \times a \:  \times s} \:  \:  \:  \:  \:  \:  \:  \:  \: ( \: ∴ \: f \:  =  \: ma \: ; \: s \:  = d)

⇒\sf{W = m \:  \times  \frac{v - u}{t}  \times  \frac{u + v}{2}  \times t \:  \:  \:  \:  \:  \:  \:  \:  \: (∴ \: a =   \frac{v - u}{t} and \: s =  \frac{u + v}{2}  \times t}

⇒\sf{W = m \times  \frac{ {v}^{2}  -  {u}^{2} }{2} }

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The work done becomes the kinetic energy of the body. Thus, the kinetic energy of a body of mass ' m : moving with the velocity equal to 'v ' is 1 / 2 mv²

∴ \sf{KE=  \frac{1}{2} m {v}^{2} }

\sf{ \underline{ \bold{  {proved}}}}

Hope I helped!

Best regards!!

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3 years ago
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