A blue ball is thrown upward with an initial speed of 23.9 m/s, from a height of 0.8 meters above the ground. 2.9 seconds after
1 answer:
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Their velocity afterwards is 2.88 m/s east
Explanation:
We can solve this problem by using the law of conservation of momentum. In fact, for an isolated system (= no external force), the total momentum must be conserved before and after the collision. So we can write:
where: in this case:
is the mass of the first player
is the initial velocity of the first player (choosing east as positive direction)
is the mass of the second player
is the initial velocity of the second player
is their combined velocity afterwards
Solving for v, we find:
And the sign is positive, so the direction is east.
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Answer:
t_{out} = t_{in}, t_{out} =
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D /
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D /
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D /
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =