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3 years ago

7

How much must a woman weigh ( force) if the pressure she exerts while standing on one foot has an area of 0.6m squared exerts a

pressure of 16 Pa

1 answer:

lions [1.4K]3 years ago

4 0

Answer:

W = 9.6 N

Explanation:

Given that,

Area on 1 foot, A = 0.6 m²

Pressure, P = 16 Pa

The pressure is given by force acting per unit area. So,

$P=\dfrac{F}{A}\\\\P=\dfrac{W}{A}\\\\W=16\times 0.6\\\\W=9.6\ N$

So, the required weight is 9.6 N.

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If a force of 32000N exerted pressure of 160N/m² , find the area on which the force acts.

Alex17521 [72]

Answer: 200m^2

Explanation:

160N=32000N/x

x*160N=32000

x=200m^2

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1 year ago

What is it called when velocity changes over time

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Well,

When an object's velocity changes, we call it acceleration.

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4 years ago

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Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8

umka2103 [35]

Answer:

a) $F=2.048\times 10^{-7}\ N$

b) $a=0.1138\ m.s^{-2}$

Explanation:

Given:

mass of raindrops, $m=1.8\times 10^{-6}\ kg$

charge on the raindrops, $q=+21\times 10^{-12}\ C$

horizontal distance between the raindrops, $r=0.0044\ m$

A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

$F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}$

we have:

$\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}$

<em>Now force:</em>

$F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}$

$F=2.048\times 10^{-7}\ N$

B)

<u>Now the acceleration on the raindrops due to the electrostatic force:</u>

$a=\frac{F}{m}$

$a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}$

$a=0.1138\ m.s^{-2}$

7 0

3 years ago

Read 2 more answers

What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

3 0

3 years ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

levacccp [35]

Answer:

The minimum speed must the car must be 13.13 m/s.

Explanation:

The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.

We know that, mg be the weight of car and rider, which is equal to the centripetal force.

$mg = \dfrac{mv^2}{r}\\v = \sqrt{rg}\\v = \sqrt{17.6\times 9.8}\\v = 13.13\ m/s\\$

So, the minimum speed must the car must be 13.13 m/s.

5 0

3 years ago