A skateboarder traveling at 4.45 m/s can be stopped by a strong force in 1.82 s and by a weak force in 5.34 s.
1 answer:
1) Impulse:
2) Strong force: -131.1 N, weak force: -44.7 N
Explanation:
1)
The impulse exerted on an object is equal to the change in momentum of the object itself.
Mathematically:
where
m is the mass of the object
u is the initial velocity
v is the final velocity
For the skateboarder in this problem, we have:
m = 53.6 kg
u = 4.45 m/s
v = 0 (it comes to a stop)
Therefore, the impulse is
Where the negative sign indicates that the direction is opposite to the motion of the object.
2)
The impulse is also equal to the product between the force applied and the duration of the collision:
where
I is the impulse
F is the average force
is the time during which the force is applied
The strong force is applied in a time of
Therefore this force is
The weak force is applied in a time of
So this force is
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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} =