V^2-u^2=2as
v=final velocity=unkown
u=initial velocity=0 m/s, because freely falling
a=acceleration due to gravity=9.8 m/s^2
s=distance (here height) traveled=4.5m
therefore the final velocity,
v^2=2*9.8*4.5
v=<span>9.39m/s</span>
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).
Find the velocity of the object after one second.
v = vo + at
v = (0 m/s) + (9.8 m/s^2)(1 s)
v = 9.8 m/s
Now, using that, you can find the displacement in that one second between 1 and 2.
d = vot + (1/2)at^2
d = (9.8 m/s)(1 s) + (1/2)(9.8 m/s^2)(1 s)^2
d = 14.7 m
<span>They can help you find services in your community.</span>
