Atmospheric conditions near the Gulf of Mexico are such that a cloud at an altitude of 1.90 km contains 3.20 ✕ 107 kg of water vapor. How long (in years) would it take for a 2.21 kW pump to raise this amount of water (at a constant speed) from Earth's surface to the altitude of the cloud?

t = 8.55 years

Explanation:

The formula for power is given as:

Power = Work/Time

but, the work here is equal to the increase in potential energy of the water. Therefore,

Power = P.E/Time

P = mgh/t

t = mgh/P

where,

t = time required = ?

P = Power of pump = 2.21 KW = 2210 W

m = mass of water = 3.2 x 10⁷ kg

h = height = 1.9 km = 1900 m

g = 9.8 m/s²

Therefore,

t = (3.2 x 10⁷ kg)(9.8 m/s²)(1900 m)/(2210 W)

t = (2.69 x 10⁸ s)(1 h/3600 s)(1 day/24 h)(1 year/365 days)

t = 8.55 years

powerful computers now, not then

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.