Answer:
w = 5832.372 Joules
Explanation:
Mass of water, m = 20 kg
The water was pulled up to a height of 35 meters, i.e. h = 35 m
It takes 14 minutes to pull up the water through the height, 35 m
speed = distance/ time = 35/14 = 2.5 m/min
The bucket's height, y = speed * time = 2.5t meters
6 kg of water drips out of the bucket throughout the 14 minutes
The rate at which the water drips drips out = (6/14) = 0.4286 kg/min
Mass of water that drips out in time, t = 0.4286t kg
The mass of water remaining = (20 - 0.4286t) kg
Change in Workdone, Δw = mgΔy
Δy = 2.5 Δt
Δw = mg * 2.5 Δt
dw = (20 - 0.4286t)g2.5 dt
integrating both sides
dw = (50g - 1.07gt)dt
where b = 0, a = 14
w = 50gt - 1.07g(t²)/2 g = 9.8 m/s²
w = 490t - 5.243t²
w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)
w = 6860 - 1027.628
w = 5832.372 Joules
Answer: D
If the fog disappears when the Sun comes out, then this is an example of condensation because:
the Sun actually dries up the fog, and it makes it into higher clouds.
Hope this helps you!
Answer:
1. 0.574 kJ/kg
2. 315.7 MW
Explanation:
1. The mechanical energy per unit mass of the river is given by:


Where:
Ek is the kinetic energy
Ep is the potential energy
v is the speed of the river = 3 m/s
g is the gravity = 9.81 m/s²
h is the height = 58 m

Hence, the total mechanical energy of the river is 0.574 kJ/kg.
2. The power generation potential on the river is:

Therefore, the power generation potential of the entire river is 315.7 MW.
I hope it helps you!
Answer:
4.7 s
Explanation:
The complete question is presented in the attached image to this solution.
v(t) = 61 - 61e⁻⁰•²⁶ᵗ
At what time will v(t) = 43 m/s?
We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.
43 = 61 - 61e⁻⁰•²⁶ᵗ
- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18
e⁻⁰•²⁶ᵗ = (18/61) = 0.2951
In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205
-0.26t = -1.2205
t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.
Hope this Helps!!!