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3 years ago

100 POINTS!!!! Use complete sentences to explain how breathing is affected in high altitudes.

2 answers:

6 0

At high altitudes, the air pressure is lower and it is more difficult for us to breathe because we cannot take in enough oxygen.

andre [41]3 years ago

4 0

Answer:

higher altitude make it harder for people to breathe

Explanation:

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Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

nirvana33 [79]

Answer:

1.327363 m/s

0.00090243 m

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of flea

Energy

$E=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 3.7\times 10^{-4}=2.1\times 10^{-4}(v^2-0)\\\Rightarrow v=\sqrt{\frac{3.7\times 10^{-4}}{2.1\times 10^{-4}}}\\\Rightarrow v=1.32736\ m/s$

The velocity of the flea when leaving the ground is 1.327363 m/s

$W=F\times s\\\Rightarrow s=\frac{W}{F}\\\Rightarrow s=\frac{3.7\times 10^{-4}}{0.41}\\\Rightarrow s=0.00090243\ m$

The flea will travel 0.00090243 m upward

8 0

3 years ago

As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax

AysviL [449]

Answer:

$I=2.766\ kg.m^2$

Explanation:

We have:

diameter of the wheel, $d=0.88\ m$

weight of the wheel, $w_w=280\ N$

mass of hanging object to the wheel, $m_o=6.32\ kg$

speed of the hanging mass after the descend, $v_o=4\ m.s^{-1}$

height of descend, $h=2.5\ m$

(a)

moment of inertia of wheel about its central axis:

$I=\frac{1}{2} m.r^2$

$I=\frac{1}{2} \frac{w_w}{g}.r^2$

$I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2$

$I=2.766\ kg.m^2$

3 0

4 years ago

7. A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g o

dolphi86 [110]

Answer:

37.34372 kg

Explanation:

m = Mass

$\Delta T$ = Change in temperature

1 denotes water

2 denotes copper

c = Heat capacity

Heat is given by

$Q=mc\Delta T$

In this case the heat transfer will be equal

$m_1c_1\Delta T_1=m_2c_2\Delta T_2\\\Rightarrow m_2=\frac{m_1c_1\Delta T_1}{c_2\Delta T_2}\\\Rightarrow m_2=\frac{95.7\times 4.18(24.2-22.7)}{0.39(65.4-24.2)}\\\Rightarrow m_2=37.34372\ kg$

Mass of copper block is 37.34372 kg

5 0

3 years ago

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.

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

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