Answer:
The latent heat of vaporization of water is 2.4 kJ/g
Explanation:
The given readings are;
The first (mass) balance reading (of the water) in grams, m₁ = 581 g
The second (mass) balance reading (of the water) in grams, m₂ = 526 g
The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ
The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ
The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature
Based on the measurements, we have;
The latent heat of vaporization = ΔQ/Δm
∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g
The latent heat of vaporization of water = 2.4 kJ/g
A.The swimmer pushes the water
C. the walls force against the ball
Answer:
170N
Explanation:
First add 530N to 150N and you get 680N, then add 400N to 450N and get 850N. So subtract 850N by 680N and you get 170N
Answer:
600 Joules
Explanation:
Using the formula F*d*cosФ. Assuming the Ф is parallel to the motion. The work done is 600 Joules.
Answer: 20.4m
Explanation:
Mass = 0.145kg
Initial velocity, Vi =20m/s
Initial kinetic energy K =1/2mv^2
Initial potential energy Ui = mgx = 0joules
: From conservation of energy,
Uf + Kf = Ui + Ki ( where f represent (final) )
Thus
mgXf + 0 = 0+1/2 mv^2
Xf = Vi^2/ 2g
= (20m/s) ^2/ 2(9.81m/s)^2
=20.4m
