Answer:
[tex] m_{fresh}= 1000 \frac{Kg}{m^3} * 2.5x10^{-4} m^3 = 0.25 Kg[/tex]
And we can do a similar procedure for the sea water:
[tex] m_{sea}= \rho_{sea} V_{sea} [/tex]
And after convert the volume to m^3 we got:
[tex] m_{sea}= 1030 \frac{Kg}{m^3} * 1x10^{-4} m^3 = 0.103 Kg[/tex]
And then the density for the mixture would be given by:
[tex] \rho_{mixture}= \frac{m_{fresh} +m_{sea}}{v_{fresh} +v_{sea}}[/tex]
And replacing we got:
[tex] \rho_{mixt}= \frac{0.25 +0.103 Kg}{2.5x10^{-4} m^3 +1x10^{-4} m^3} = 1008.571 \frac{Kg}{m^3}[/tex]
Step-by-step explanation:
For this case we can begin calculating the mass for each type of water:
[tex] m_{fresh}= \rho_{fresh} V_{fresh} [/tex]
And after convert the volume to m^3 we got:
[tex] m_{fresh}= 1000 \frac{Kg}{m^3} * 2.5x10^{-4} m^3 = 0.25 Kg[/tex]
And we can do a similar procedure for the sea water:
[tex] m_{sea}= \rho_{sea} V_{sea} [/tex]
And after convert the volume to m^3 we got:
[tex] m_{sea}= 1030 \frac{Kg}{m^3} * 1x10^{-4} m^3 = 0.103 Kg[/tex]
And then the density for the mixture would be given by:
[tex] \rho_{mixture}= \frac{m_{fresh} +m_{sea}}{v_{fresh} +v_{sea}}[/tex]
And replacing we got:
[tex] \rho_{mixt}= \frac{0.25 +0.103 Kg}{2.5x10^{-4} m^3 +1x10^{-4} m^3} = 1008.571 \frac{Kg}{m^3}[/tex]
