Alarge, cylindrical water tank with diameter 2.40 m is on a platform 2.00 m above the ground. the vertical tank is open to the air and the depth of the water in the tank is 2.00 m. there is a hole with diameter 0.520 cm in the side of the tank just above the bottom of the tank. the hole is plugged with a cork. you remove the cork and collect in a bucket the water that flows out the hole. a) when 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank?

b) how long does it take you to collect 1.00 gal of water in the bucket?

a.  h₁ =0.0008m = 0.8mm is the height of water that flow out

the change in height of water in the tank = h - h₁ = 2 - 0.0008 = 1.9992m

b.    t =31.93 sec

Explanation:

a) When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank?

given:,

cylindrical water tank with diameter, D - 2.40 m

the tank is 2.00 m above the ground.

depth of the water in the tank is 2.00 m.

diameter of hole, d - 0.520 cm

firstly, we need to calculate the volume of water in the tank:

Volume = πr²h

              =(pi * D * D)/4  * height of water

              =(3.142 * 2.4 * 2.4)/4  * 2 = 4.52*2 = 9.05m³

1.0 gal of water is equivalent to = 0.0038m³

the volume of 1gal is 0.0038 = A *h

the area of the tank is calculated above as 4.52m²

therefore, 0.0038 = 4.52 *h

                    h₁ =0.0008m = 0.8mm is the height of water that flow out

the change in height of water in the tank = h - h₁ = 2 - 0.0008 = 1.9992m

b) How long does it take you to collect 1.00 gal of water in the bucket?

1.0 gal of water is equivalent to = 0.0038m³

to calculate for Volume flow, Q for a draining tank

       Q = Cd * A *

where Cd is a discharge coefficient, and is given by 0.9 for water

A is the area of the small hole = (pi * D * D)/4 = (pi * 0.0052 * 0.0052)/4

          A = 0.0000212m²

H= height of the hole from the tank water level= 2m - 0.0052 = 1.9948m

g = 9.8m/s2

 Q = 0.9 * 0.0000212 *

 Q = 0.0000191 * 6.253 = 0.000119 m³/s

 Q = V/t

 Qt = V

   t = V/Q = 0.0038m³/0.000119 m³/s

   t =31.93 sec

number of electrons for neutral cl

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.