The atoms furthest from the nucleus
During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to

where
m is the mass of the water

is the latent heat of evaporation.
For water, the latent heat of evaporation is

, while the mass of the water is

so, the amount of heat released in the process is
The net force applied to the object equals the mass of the object multiplied by the amount of its acceleration." The net force acting on the soccer ball is equal to the mass of the soccer ball multiplied by its change in velocity each second (its acceleration).
Answer:
it would be c i just had it
Explanation:
welcome................
Answer:
The force on one side of the plate is 3093529.3 N.
Explanation:
Given that,
Side of square plate = 9 m
Angle = 60°
Water weight density = 9800 N/m³
Length of small strip is


The area of strip is

We need to calculate the force on one side of the plate
Using formula of pressure


On integrating




Hence, The force on one side of the plate is 3093529.3 N.