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ExtremeBDS [4]
3 years ago
10

Help ASAP!

Physics
1 answer:
mars1129 [50]3 years ago
5 0

Answer:

I hope the answer helps you

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Water waves approach an underwater "shelf" where the velocity changes from 2.8 m/s to 2.1 m/s. If the incident wave crests make
otez555 [7]

Answer:

24°

Explanation:

sin(34°)/sin(x)=v2/v1

x=arcsin(2,1*sin(34°)/2,8)=24°

7 0
3 years ago
Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water
s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

a

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is \frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5

b

The number of moles of gas that must be released is  n= 0.3538\ mols

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

 This means that the pressure at the depth of the surface would be

                P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm

                      = 2.5 atm

The ideal gas equation is mathematically represented as

                PV = nRT

Where P is pressure at the surface

           V is the volume

            R is the gas constant  = 8.314 J/mol. K

making n the subject we have

        n = \frac{PV}{RT}

 Considering at the surface of the water the number of moles at the surface would be

               n_s = \frac{P_sV}{RT}

Substituting 1 atm = 101325 N/m^2 for P_s ,6L = 6*10^{-3}m^3 for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

              n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}

                   = 0.2359 mol  

To obtain the number of moles at the depth of the water we use

                n_d  = \frac{P_d V}{RT}

Where P_d \ and \ n_d \ are pressure and no of moles at the depth of the water

        Substituting values we have

              n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}

                  = 0.5897 mol

Now to obtain the number of moles released we have

             n =  n_d - n_s

               = 0.5897mol  - 0.2359mol

              =0.3538 \ mol

     The molar concentration at the surface  of water is

                [\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}

                                =39.31mol/m^3

    The molar concentration at the depth  of water is

           [\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}

                      = 98.28 mol/m^3

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

         \frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5

                   

                     

                     

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A motor does a total of 480 joules of work in 5.0 seconds to lift a 12-kilogram block to the top of a rampThe average power deve
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Answer:

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What is hard water and what is soft water
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Water that is "hard" <span>contains a certain amount of dissolved minerals (like calcium and magnesium). 
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What is the difference in sodium and chlorine
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sodium is a metal in group 1

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3 years ago
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