Answers:
a) [tex]1.307(10)^{21} kg[/tex]
b) [tex]1051.359 J[/tex]
c) [tex]5.777 m/s[/tex]
Explanation:
Density [tex]\rho[/tex] is defined as a relation between mass [tex]m[/tex] and volume [tex]V[/tex]:
[tex]\rho=\frac{m}{V}[/tex] (1)
Where:
[tex]\rho=2750 kg/m^{3}[/tex] is the average density of the continent
[tex]m[/tex] is the mass of the continent
[tex]V[/tex] is the volume of the continent, which can be estimated is we assume it as a a slab of rock 4050 km on a side and 29 km deep:
[tex]V=(length)(width)(depth)=(4050 km)(4050 km)(29 km)=475,672,500 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=4.756(10)^{17} m^{3}[/tex]
Finding the mass:
[tex]m=\rho V[/tex] (2)
[tex]m=(2750 kg/m^{3})(4.756(10)^{17} m^{3})[/tex] (3)
[tex]m=1.307(10)^{21} kg[/tex] (4) This is the mass of the continent
Kinetic energy [tex]K[/tex] is given by the following equation:
[tex]K=\frac{1}{2}mv^{2}[/tex] (5)
Where:
[tex]m=1.307(10)^{21} kg[/tex] is the mass of the continent
[tex]v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1.2683(10)^{-9} m/s[/tex] is the velocity of the continent
[tex]K=\frac{1}{2}(1.307(10)^{21} kg)(1.2683(10)^{-9} m/s)^{2}[/tex] (6)
[tex]K=1051.359 J[/tex] (7) This is the kinetic energy of the continent
If we have a jogger with mass [tex]m=63 kg[/tex] and the same kinetic energy as that of the continent [tex]653.5 J[/tex], we can find its velocity by isolating [tex]v[/tex] from (5):
[tex]v=\sqrt{\frac{2 K}{m}}[/tex] (6)
[tex]v=\sqrt{\frac{2 (1051.359 J)}{63 kg}}[/tex]
Finally:
[tex]v=5.777 m/s[/tex] This is the speed of the jogger
Answer:
a) [tex]1.31x10^{21} kg[/tex].
b) [tex]1.51x10^{3} J[/tex]
c) [tex]6.92\frac{m}{s}[/tex]
Explanation:
In first place, to calculate the mass of the continent we have to use the density definition because it includes volume (V), density (p) and mass (m), and the problem give us all values needed. The equation to solve is [tex]\rho=\frac{m}{V}[/tex]; we have to isolate m and first calculate the volume.
[tex]V = A.h =(4050km)(4050km)(29km) = 4.76x10^{8} km^{3}[/tex]; but we need to use volume in meters.
[tex]1km = 1000m\\1km^{3} =(1000m)^{3}=1x10^{9} m^{3}[/tex]
Then, [tex]V=4.76x10^{8} km^{3} \frac{1x10^{9}m^{3} }{1km^{3} } = 4.76x10^{17}m^{3}[/tex]
[tex]m = \rho.V=2750\frac{kg}{m^{3} } (4.76x10^{17}m^{3} )=1.31x10^{21} kg[/tex].
Then, to calculate the kinetic energy, we recur to the definition expressed in this equation: [tex]K=\frac{1}{2} mv^{2}[/tex]. But, we first have to transform the continent speed into meter and seconds, because it's in centimeter and years.
We know that 1 meter is equivalent 100 centimetres, and 1 year is equivalent to [tex]3.15x10^{7} sec[/tex]. So, we transform:
[tex]v=4.8\frac{cm}{year} .\frac{1year}{3.15x10^{7}sec }.\frac{1m}{100cm} = 1.52x10^{-9} \frac{m}{s}[/tex].
[tex]K= \frac{1}{2} (1.31x10^{21}kg)(1.52x10^{-9}\frac{m}{s} )^{2} = 1.51x10^{3} J[/tex]
We use the same definition and isolate the speed in the equation:
[tex]K=\frac{1}{2} mv^{2} \\v=\sqrt{\frac{2K}{m} }=\sqrt{\frac{2(1.51x10^{3}J) }{63kg} } =6.92\frac{m}{s}[/tex]
Therefore, the speed of the jogger is [tex]6.92\frac{m}{s}[/tex]
