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9966 [12]
3 years ago
9

Atmosphere in a sealed space craft contains

Physics
1 answer:
Alekssandra [29.7K]3 years ago
8 0
The best and most correct answer among the choices provided by your question is the fourth choice or letter D.

<span>Atmosphere in a sealed space craft contains </span><span>pressurized atmospheric air available normally on earth.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
You might be interested in
The noble gases neon (atomic mass 20.1797 u) and krypton (atomic mass 83.798 u) are accidentally mixed in a vessel that has a te
vagabundo [1.1K]

Answer:

(a) Kav Ne = Kav Kr = 7.29x10⁻²¹J

(b) v(rms) Ne= 659.6m/s and v(rms) Kr= 323.7m/s

Explanation:

(a) According to the kinetic theory of gases the average kinetic energy of the gases can be calculated by:

K_{av} = \frac{3}{2}kT (1)        

<em>where K_{av}: is the kinetic energy, k: Boltzmann constant = 1.38x10⁻²³J/K, and T: is the temperature </em>

<u>From equation (1), we can calculate the</u><u> average kinetic energies for the krypton and the neon:</u>

K_{av} = \frac{3}{2} (1.38\cdot 10^{-23} \frac{J}{K})(352.2K) = 7.29\cdot 10^{-21}J  

(b) The rms speeds of the gases can be calculated by:

K_{av} = \frac{1}{2}mv_{rms}^{2} \rightarrow v_{rms} = \sqrt \frac{2K_{av}}{m}  

<em>where m: is the mass of the gases and v_{rms}: is the root mean square speed of the gases</em>

For the neon:

v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{20.1797 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 659.6 \frac{m}{s}          

For the krypton:

v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{83.798 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 323.7 \frac{m}{s}  

Have a nice day!

3 0
3 years ago
What is the magnitude of the electric field at a point that is 0.5m to the right of a small sphere with a net charge of -6.0x10-
zysi [14]

Answer:

lil durk a goat

Explanation:

8 0
2 years ago
A graphic designers is not a visual artist.<br> True <br> False
Colt1911 [192]

This answer is false. A visual artist is defined as someone involved in the arts of painting, sculpting, photography, etc, as opposed to music, drama, and literature. A visual artist is in the form of making somthing more visable. So, a graphic designer is a visual artist.

5 0
3 years ago
Read 2 more answers
Air in human lungs has a temperature of 37.0°C and a saturation vapor density of 44.0 g/m³.
ioda

Answer:

(a). The maximum loss of water vapor by the person is 8.8\times10^{-2}\ g

(b). The partial pressure of water vapor is 6.29\times10^{3}\ N/m^2

Explanation:

Given that,

Temperature = 37.0°C

Volume of air = 2.00 L

Density of vapor = 44.0 g/m³

We need to calculate the maximum loss of water vapor by the person

Using formula of density

m=\rho_{air}\times V_{air}

Put the value into the formula

m=44.0\times2.00\times10^{-3}

m=0.088\ g

m=8.8\times10^{-2}\ g

(b). We need to calculate the partial pressure of water vapor

Using formula of pressure

PV=nRT

P=\dfrac{nRT}{V}

P=\dfrac{\rho\times R\times T}{M}

Put the value into the formula

P=\dfrac{44\times8.314\times(37+273)}{18.01528}

P=6294.82\ N/m^2

P=6.29\times10^{3}\ N/m^2

Hence, (a). The maximum loss of water vapor by the person is 8.8\times10^{-2}\ g

(b). The partial pressure of water vapor is 6.29\times10^{3}\ N/m^2

3 0
3 years ago
A thin film of cooking oil (n=1.43) is spread on a puddle of water (n=1.34). What is the minimum thickness Dmin of the oil that
Morgarella [4.7K]

Complete Question

A thin film of cooking oil (n=1.43) is spread on a puddle of water (n=1.34). What is the minimum thickness Dmin of the oil that will strongly reflect blue light having a wavelength in air of 451 nm, at normal incidence? Dmin= nm What are the next three thicknesses that will also strongly reflect blue light of the same wavelength, at normal incidence?

a)   158 nm, 237 nm, and 315 nm

b)   237 nm, 394 nm, and 552 nm

c)   473 nm, 788 nm, and 1100 nm

d)  361 nm, 602 nm, and 842 nm

e)  315 nm, 473 nm, and 631 nm

Answer:

The minimum thickness is   t= 78.8nm

The correct option is B

Explanation:

  From the question we are told that

      The refractive index of cooking oil is n _c = 1.43

       The refractive index of water is n_w = 1.34

      The  wavelength of reflection is is  \lambda _ r = 451nm

The formula for the thickness of the oil film is mathematically represented as

        2 n t = (m + \frac{1}{2} ) \lambda

Where n is the refractive index of oil

            m is the integer number of fringe

            t is the thickness

   for a minimum reflection  m= 0

Now making t the subject of the formula

         t =  \frac{(m + \frac{1}{2} \lambda ) }{2 n}

Substituting value

        t = \frac{(0 + 0.5) * 451 *10^{-9}}{2 * 1.43}

          t= 78.8nm

For the next thickness m = 1

   so we have

        t_1 = \frac{(1 + 0.5 ) * 481}{2 * 1.43}

            = 237nm

For the next thickness m = 2

     so we have that  

        t_2 = \frac{(2 +0.5) *451 *10^{-9}}{2 * 1.43}

             = 394nm

For the next thickness m = 3

     so we have that

         t_2 = \frac{(3 +0.5) *451 *10^{-9}}{2 * 1.43}

             = 552nm

6 0
3 years ago
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