Question

2.05 × 108 kg of water is pumped up through a vertical height of 500 m.

Answer 1

The speed at which the water leaves the pipe is 98.34 m/s.

Step-by-step explanation:

The speed at which water leaves the pipe can be determined using the principle of conservation of energy. The energy gained by the water as it is pumped up through the vertical height is equal to the energy it loses when it reaches the other end of the pipe and is open to the air.

First, we need to calculate the gravitational potential energy gained by the water. The mass of the water is given as 2.05 × 108 kg. The potential energy can be calculated using the formula: PE = m × g × h. Substituting the values, we get: PE = 2.05 × 108 kg × 9.8 m/s2 × 500 m = 9.95 × 1011 J.

Next, we can calculate the kinetic energy of the water using the formula: KE = 1/2 × m × v2. The velocity of the water can be assumed to be zero at the top of the pipe, so the kinetic energy is zero. Therefore, the total mechanical energy of the water is equal to the potential energy: E = PE = 9.95 × 1011 J.

The speed at which the water leaves the pipe can be calculated using the formula: E = 1/2 × m × v2. Rearranging the equation, we get: v = √(2E / m). Substituting the values, we find: v = √(2 × 9.95 × 1011 J / 2.05 × 108 kg) = √(9.68 × 103) = 98.34 m/s.

Answer 2

1 x 10¹²J

Step-by-step explanation:

Given parameters:

Mass of the water pumped = 2.05 x 10⁸kg

Height = 500m

Unknown:

Potential energy = ?

Solution:

The potential energy amassed by water at this height is expressed as;

P.E = mgH

m is the mass

g is the acceleration due to gravity

H is the height

Insert the parameters and solve;

P.E = 2.05 x 10⁸ x 9.8 x 500 = 1 x 10¹²J