When the height of the cylinder increased by a factor of 5, from 100 m to 500 m, what happened to the cylinder’s gravitational p
2 answers:
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To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,
Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that
Therefore the final kinetic energy is 3600MJ
Answer:
0.423m
Explanation:
Conversion to metric unit
d = 4.8 cm = 0.048m
Let water density be
Let gravitational acceleration g = 9.8 m/s2
Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.
Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:
Where N is the spring force,
is the buoyancy force, which equals to the weight
of the water displaced by the submerged portion of the cylinder, which is the product of water density
, submerged volume
and gravitational constant g. W = mg is the weight of the metal cylinder.
The submerged volume would be the product of cross-section area and the submerged length x
Plug that into our force equation and we have