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.

3 0

3 years ago

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Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cook

jeka57 [31]

Answer: a) Mr = 2.4×10^-4kg/s

V = 34.42m/a

b) E = 173J

Ø = 2693.1J

c) Er = 0.64J/s

Explanation: Please find the attached file for the solution

3 0

3 years ago

A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t

vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

$\mu =7.20 g/m = 0.0072 kg/m$ is the mass linear density

Solving the equation,

$f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz$

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

$f=\frac{v}{2L}$

where

v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

$L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m$

4 0

4 years ago

An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V/√g.l where V is a

prohojiy [21]

Answer:

1.24611

Explanation:

V = Velocity = 10 ft/s

L = Length = 2 ft

g = Acceleration due to gravity = 32.2 ft/s²

Froude number is given by

$Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10}{\sqrt{32.2\times 2}}\\\Rightarrow Fr=1.24611$

Converting to SI units

$10\ ft/s=10\times \dfrac{1}{3.281}$

$32.2\ ft/s^2=32.2\times \dfrac{1}{3.281}$

$2\ ft=2\times \dfrac{1}{3.281}$

$Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10\times \dfrac{1}{3.281}}{\sqrt{32.2\times \dfrac{1}{3.281}\times 2\times \dfrac{1}{3.281}}}\\\Rightarrow Fr=1.24611$

The Froude number is 1.24611

The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.

7 0

3 years ago