A dam 143 m above a hydroelectric generator holds 1.6 x 1015 L of water. If the generator is taken to be at a height of 0 m, how much potential energy is stored in the dam?
I assume you made a typo and mean 1.6 * 10^15 Liters
1 liter water = 1 kg
a pint is a pound the world around :)
m g h = 1.6*10^15 kg * 9.81 m/s^2 * 143 m
= 2245 * 10^15
= 2.245 * 10^18 Joules
PE=mgh
=1.6*10^15*9.8*143
=2242.24*10^15
=2.242*10^18
To calculate the potential energy stored in the dam, we can use the formula:
Potential energy = mass * gravity * height
In this case, the mass of the water is given as 1.6 x 10^15 L. However, we need to convert this volume to mass by multiplying it by the density of water. The density of water is approximately 1 g/cm^3, or 1000 kg/m^3.
To convert the volume from liters to cubic meters, we divide by 1000.
So, the mass of the water is:
1.6 x 10^15 L * (1 m^3 / 1000 L) * (1000 kg / 1 m^3) = 1.6 x 10^18 kg
The acceleration due to gravity, denoted by "g," is approximately 9.8 m/s^2.
The height is given as 143 m.
Now, we can substitute these values into the formula to calculate the potential energy:
Potential energy = (1.6 x 10^18 kg) * (9.8 m/s^2) * (143 m)
Therefore, the potential energy stored in the dam is:
Potential energy = 2.3 x 10^21 Joules