A certain rain cloud at an altitude of 1.80 km contains 3.20 107 kg of water vapor. How long would it take for a 2.93-kW pump to raise the amount of water from Earth's surface to the cloud's position?

Respuesta :

Answer:

192851877 s or 6.1 years

Explanation:

1.8 km = 1800 m

2.93 kW = 2930 W

First we calculate the potential energy needed to pump [tex]3.2\times10^7kg[/tex] of water to 1800 m high. Suppose g = 9.81 m/s2:

[tex]E = mgh = 3.2*10^7*1800*9.81 = 5.65 \times 10^{11} J[/tex]

Then we can calculate the time it takes for a pump with power of 2930 W to perform that amount of energy

[tex]t = E/P = 5.65 \times 10^{11}/2930 = 192851877 s = 192851877.1/(60*60) = 53570 hours = 53570 / 24 = 2232 days = 2232 / 365.25 = 6.1 years[/tex]

Answer:

Explanation:

The potential energy contained in the mass of water by using the equation for gravitational potential energy, G.P.E = mass x height x acc. due to gravity.

E = 3.2 x 10^7 × 1.80 x 1000 × 9.81

= 5.65 × 10^11 J

Power = Work done/time taken

2.93 × 10^3 = 5.65 × 10^11/t

T = 5.488 10^11/2700

= 1.93 × 10^8 s.

Time taken, t = 1.93 × 10^8 s

= 6.12 years.