Answer:
[tex]W=2.77J[/tex]
Explanation:
The amount of work required will be the energy needed to elevate the volume of blood mentioned (that is, the mass of blood mentioned) by the distance mentioned. This energy is the gravitational potential energy, so we have:
[tex]W=E=U_g=mgh=\rho Vgh[/tex]
We only have to be careful in writing everything in S.I., and for this we will use conversion factors, which are equal to 1 (and only change our units):
[tex]\rho=1.06 g/cm^3=\frac{1.06 g}{cm^3}(\frac{1kg}{1000g})(\frac{100cm}{1m})^3=1060kg/m^3[/tex]
[tex]V=0.816L=816mL=816mL(\frac{1cm^3}{1mL})(\frac{1m}{100cm})^3=0.000816m^3[/tex]
(Notice that every term between parenthesis is just equal to 1, and we place the units so the ones we don't want cancel out).
[tex]W=\rho Vgh=(1060kg/m^3)(0.000816m^3)(9.8m/s^2)(0.327m)=2.77J[/tex]
Answer:
The work done by heart is 2.77 J
Explanation:
Given data:
Volume of blood is, [tex]V = 0.816 \;\rm L=0.816 \times 10^{-3} \;\rm m^{3}[/tex]
Vertical distance travelled by blood is, [tex]h = 0.327 \;\rm m[/tex].
Density of blood is, [tex]\rho =1.06 \;\rm g/cm^{3} = 1.06 \times 1000 =1060 \;\rm kg/m^{3}[/tex].
The work done by the heart to pump the blood is,
[tex]W = \rho \times V \times gh[/tex]
here, g is gravitational acceleration.
[tex]W = 1060 \times 0.816 \times 10^{-3} \times 9.8 \times 0.327\\W=2.77 \;\rm J[/tex]
Thus, work done by heart is 2.77 J.
For more details, refer the link:
https://brainly.com/question/22599382?referrer=searchResults
