As ocean waves get closer to the shore they slow down. This slowing causes them to bend and come in close to parallel with the s
2 answers:
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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} =
Answer:
(a):
(b):
(c):
(d):
(e):
Explanation:
Given, the position of the particle along the x axis is
The units of terms and
should also be same as that of x, i.e., meters.
The unit of t is seconds.
(a):
Unit of
Therefore, unit of
(b):
Unit of
Therefore, unit of
(c):
The velocity v and the position x of a particle are related as
(d):
The acceleration a and the velocity v of the particle is related as
(e):
The particle attains maximum x at, let's say, , when the following two conditions are fulfilled:
Applying both these conditions,
For ,
Since, c is a positive constant therefore, for ,
Thus, particle does not reach its maximum value at
For ,
Here,
Thus, the particle reach its maximum x value at time