Answer:
The recoil speed of Astronaut A is 0.26 m/s.
Explanation:
Given that,
Mass of astronaut A, 
Mass of astronaut B, 
Astronaut A pushes B away, with B attaining a final speed of 0.4, 
We need to find the recoil speed of astronaut A. The momentum remains conserved here. Using the law of conservation of linear momentum as :

So, the recoil speed of Astronaut A is 0.26 m/s.
For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:

which becomes

where

is the force on the first piston

is the area of the first piston

is the force on the second piston

is the area of the second piston
If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
Current will be

now just pluf in the values and Voila..
Answer:
V=4.7m/s
Explanations:
Let Ma mass of cat A=7kg
Va velocity of cat A=7m/s
Mb mass of cat b=6.1kg
VB velocity of cat b=2m/s
From conservation of linear momentum
MaVa+MbVb=(Ma+Mb)V
7*7+6.1*2=(7+6.1)V
61.2=13.1V
V=4.7m/s
Answer:
2.47 m
Explanation:
Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.
The horizontal velocity of the ball is constant:

and the time taken to cover the horizontal distance d is

So this is the time the ball takes to reach the horizontal position of the crossbar.
The vertical position of the ball at time t is given by

where
is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

The height of the crossbar is h = 3.05 m, so the ball passes

above the crossbar.