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3 years ago

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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A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

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$x = -19 + t - 3t^3$

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Part a)

differentiate x and y two times with respect to time to find the acceleration

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$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

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$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

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$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

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$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

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