Init.
2 answers:
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Answer:
Explanation:
From the position coordinates given , it appears that the ball moves simultaneously along x and y direction.
Displacement along x direction in one second = 4.4 - 1.8 = 2.6 m
So velocity along x direction V_x =
Similarly velocity along y direction V_y(1) =
In the next phase velocity changes both in x and y direction.
velocity in x - direction V_x(2) = [tex]\frac{2}{s}[/tex
Velocity in Y- direction V_y(2) = [tex]\frac{3.1}{s}[/tex
Acceleration in x direction = change of velocity in x direction
= ( 2 - 2.6 ) = -.6 m s⁻²
Acceleration in y direction = ( 3.1 - 2.6 ) = 0.5 m s⁻²
Total acceleration =\sqrt{( -.6 )² + ( .5 )²}
= .78 ms⁻²
Answer:
13.7m
Explanation:
Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.
After the push
Where is the mass of the astronaut,
is the mass of the satellite,
is the speed of the satellite. We can calculate the speed
of the astronaut:
So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be
d = vt = 1.83 * 7.5 = 13.7 m