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3 years ago

Why don't you feel the pressure exerted by the atmosphere?

Physics

2 answers:

mariarad [96]3 years ago

8 0

Answer:

the air within our bodies (in our lungs and stomachs, for example) is exerting the same pressure outwards so there's no pressure difference and no need for us to exert any effort.

Explanation:

Galina-37 [17]3 years ago

6 0

Because in our body blood also exerts pressure that's why we don't feel atmospheric pressure.

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3 years ago

Electrons of wavelength 1.5 nm are emitted from a material when light of wavelength 350nm is incident on it. What is the work fu

krok68 [10]

Answer:

3.5434 eV

Explanation:

For a particle with kinetic energy E and mass m , the wavelength associated is given by the following relation,

$\lambda=\frac{h}{\sqrt{2mE} }$

E = $\frac{h^2}{2m\lambda^2}$

Putting the values we get

E = $\frac{(6.6\times(10^{-34})^2}{2\times9.1\times10^{-31\times(1.5\times10^{-9})^2}}$

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4 years ago

A 0.0223 m diameter coin rolls up a 12.0◦ inclined plane. The coin starts with an initial angular speed of 49.3 rad/s and rolls

zlopas [31]

Answer:

$h = 0.0231 m$

Explanation:

The movement of the coin is modelled after the Principle of Energy Conservation. The kinetic energy of the coin is the sum of the components associated with translation and rotation and there are no non-conservative forces. In addition, the coin starts moving at height of zero

$K_{rot} + K_{tr} = U_{g}$

$\frac{1}{2} \cdot m_{coin} \cdot \omega^{2} \cdot R^{2} + \frac{1}{4} \cdot m_{coin} \cdot R^{2} \cdot \omega^{2} = m_{coin} \cdot g \cdot h\\$

$\frac{3}{4} \cdot R^{2} \cdot \omega^{2} = g \cdot h$

The maximum vertical height is isolated in the previous equation:

$h = \frac{3 \cdot R^{2}\cdot \omega^{2}}{4\cdot g}$

$h = \frac{3 \cdot (0.01115 m)^{2} \cdot (49.3 \frac{rad}{s} )^{2}}{4 \cdot (9.807 \frac{m}{s^{2}} )} \\h = 0.0231 m$

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4 years ago