Answer:
M = 50 kg / sec * 60 sec = 3000 kg water that falls
PE = M g h initial potential energy of water
PE = 3000 kg * 9.8 m/s^2 * 20 m = 588,000 joules of energy lost
To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

Where,
Q = Heat
U = Internal Energy
By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore
Work is done ON the system
Heat is extracted FROM the system
Therefore the value of the Work done on the system is -158.0J
The woman on the platform is correct because it is the pace of the man moving on the train not walking.
Solution :
Given weight of Kathy = 82 kg
Her speed before striking the water,
= 5.50 m/s
Her speed after entering the water,
= 1.1 m/s
Time = 1.65 s
Using equation of impulse,

Here, F = the force ,
dT = time interval over which the force is applied for
= 1.65 s
dP = change in momentum
dP = m x dV
![$= m \times [V_f - V_o] $](https://tex.z-dn.net/?f=%24%3D%20m%20%5Ctimes%20%5BV_f%20-%20V_o%5D%20%24)
= 82 x (1.1 - 5.5)
= -360 kg
∴ the net force acting will be


= 218 N
Answer:
The speed of space station floor is 49.49 m/s.
Explanation:
Given that,
Mass of astronaut = 56 kg
Radius = 250 m
We need to calculate the speed of space station floor
Using centripetal force and newton's second law




Where, v = speed of space station floor
r = radius
g = acceleration due to gravity
Put the value into the formula


Hence, The speed of space station floor is 49.49 m/s.