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goblinko [34]
2 years ago
6

3) Calculate the kinetic energy of a 7 kg mass traveling at a velocity of 4 m/sec.

Physics
2 answers:
Pepsi [2]2 years ago
7 0

Answer:

56 J

Explanation:

Formula to find the kinetic energy is :

E_k = \frac{1}{2}  × m × v²

Here ,

m ⇒ mass

v ⇒  velocity

Let us solve now

 E_k = \frac{1}{2} × m × v²

     = \frac{1}{2} × 7 kg × ( 4 ms⁻¹ )²

     = \frac{1}{2} × 7 × 16

     = \frac{1}{2} × 112

     = 56 J

Hope this helps you :-)

Let me know if you have any other questions :-)

murzikaleks [220]2 years ago
4 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Let's solve ~

Given terms :

  • Mass (m) = 7 kg

  • velocity (v)= 4 m/s

The formula to find kinetic Energy is ~

\boxed{ \boxed{ \sf{ \frac{1}{2}  m{v}^{2} }}}

Now, apply the formula according to given situation

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{1}{2}  \times 7 \times ( {4)}^{2}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{1}{2}  \times 7 \times 16

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:7 \times 8

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:56 \:  \: joules

Therefore, the kinetic Energy of the car is 56 joules

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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
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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

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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

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

                   

                     

                     

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