Answer:
38.8 m
Explanation:
h = -(21 + 1.75) = - 22.75 m
g = - 9.8 m /s^2
Ux = 14.004 m/s
Uy = + 5.376 m/s
Let the ball hits the ground in time t and at a distance d from the base of hill.
Use second equation of motion
h = Uyt + 0.5 at^2
- 22.75 = 5.376 t - 0.5 x 9.8 t^2
4.9 t^2 - 5.376 t - 22.75 = 0

By solving
t = 2.77 second
So, horizontal distance
d = Ux t
d = 14.004 x 2.77 = 38.8 m
It doesn’t feel lighter it feels heavier
Answer:
The change in entropy ΔS = 0.0011 kJ/(kg·K)
Explanation:
The given information are;
The mass of water at 20.0°C = 1.0 kg
The mass of water at 80.0°C = 2.0 kg
The heat content per kg of each of the mass of water is given as follows;
The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg
The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg
Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg
The heat energy of the mixture =
1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)
∴ T = 60°C
The heat content, of the water at 60° = 251.154 kJ/kg
Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462
The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).
The distance it traveled and the time that it took to travel that distance