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valentina_108 [34]
3 years ago
14

Human body mass density?

Physics
1 answer:
navik [9.2K]3 years ago
4 0
Very slightly less than 1 g/cm^3 averaged over the entire body.

Even though it doesn't feel like it when you're in the pool, if you completely relax and let go, you do float in water.
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The instruction booklet for your pressure cooker indicates that its highest setting is 12.3 psi . you know that standard atmosph
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<span>118 C The Clausius-Clapeyron equation is useful in calculating the boiling point of a liquid at various pressures. It is: Tb = 1/(1/T0 - R ln(P/P0)/Hvap) where Tb = Temperature boiling R = Ideal Gas Constant (8.3144598 J/(K*mol) ) P = Pressure of interest Hvap = Heat of vaporization of the liquid T0, P0 = Temperature and pressure at a known point. The temperatures are absolute temperatures. We know that water boils at 100C at 14.7 psi. Yes, it's ugly to be mixing metric and imperial units like that. But since we're only interested in relative pressure differences, it's safe enough. So P0 = 14.7 P = 14.7 + 12.3 = 27 T0 = 100 + 273.15 = 373.15 And for water, the heat of vaporization per mole is 40660 J/mol Let's substitute the known values and calculate. Tb = 1/(1/T0 - R ln(P/P0)/Hvap) Tb = 1/(1/373.15 K - 8.3144598 J/(K*mol) ln(27/14.7)/40660 J/mol) Tb = 1/(0.002679887 1/K - 8.3144598 1/K ln(1.836734694)/40660) Tb = 1/(0.002679887 1/K - 8.3144598 1/K 0.607989372/40660) Tb = 1/(0.002679887 1/K - 5.055103194 1/K /40660) Tb = 1/(0.002679887 1/K - 0.000124326 1/K) Tb = 1/(0.002555561 1/K) Tb = 391.3034763 K Tb = 391.3034763 K - 273.15 Tb = 118.1534763 C Rounding to 3 significant figures gives 118 C</span>
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3 years ago
A car has 10000kg starts from rest and attains velocity of 20m/s in 5 seconds find the force
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Ewan ko po ehh sorri po’ hahaha thanks me
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The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard
gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

|\Delta E_{ij}| = |E_i-E_j|

Our states are given by

E_1 = -11.7eV

E_2 = -4.2eV

E_3 = -3.3eV

In this way the energy balance for the states would be given by,

|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\

|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV

|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

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An object is located 51 millimeter from a diverging lens. te object has a hight of 13 millimeters and the image height is 3.5 mi
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<span>An object is located 51 millimeters from a diverging lens the object has a height og 13 millimeters and the image height is 3.5 millimeters how far in front of the lens is image located?</span>
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