Most waves approach the shore at an angle. However, they bend to be nearly parallel to the shore as they approach it because when a wave reaches a beach or coastline, it releases a burst of energy that generates a current, which runs parallel to the shoreline.
- Most waves approach shore at an angle. As each one arrives, it pushes water along the shore, creating what is known as a longshore current within the surf zone.
- Waves approach the coast at an angle because of the direction of prevailing wind.
- The part of the wave in shallow water slows down, while the part of the wave in deeper water moves at the same speed.
- Thus when wave reaches a beach or coastline, it releases a burst of energy that generates a current, which runs parallel to the shoreline.
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First of all, that equation is not correct, which may be the reason
that you're having trouble assigning units to the quantities.
Power is defined as [energy / time], so [Energy] = [ power x time ],
and
[Time] = [ energy / power ].
Unit-wise, these equations are correct just as they appear here,
with no proportionality constants or conversion factors, when ...
[ Power ] = watts
[ Energy ] = joules
[ Time ] = seconds .
I'm pretty your it's magnetism though. Like magnets
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.
The work will be completely defined by
(Force) x (distance in the y-direction),
and it won't matter what route the tool follows to get anywhere.
Only the initial and final y-coordinates matter.
We know that F = - 2.85 y². (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.
From y=0 to y=2.40 is a distance of 2.40 upward.
Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.
Work = integral of (F·dy) evaluated from 0 to 2.40
= integral of (-2.85 y² dy) evaluated from 0 to 2.40
= (-2.85) · integral of (y² dy) evaluated from 0 to 2.40 .
Now, integral of (y² dy) = 1/3 y³ .
Evaluated from 0 to 2.40 , it's (1/3 · 2.40³) - (1/3 · 0³)
= 1/3 · 13.824 = 4.608 .
And the work = (-2.85) · the integral
= (-2.85) · (4.608)
= - 13.133 .
-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the -13.133 is joules.
-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40. Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up.
-- It doesn't matter whether the tool goes there along the line x=y , or
by some other route. WHATEVER the route is, the work done by ' F '
is going to total up to be -13.133 joules at the end of the day.
As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then. But that's my answer
and I'm stickin to it. If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
<span>v=u+at</span>
<span>
0=5.0283−9.8×t</span><span>
t=0.5130sec</span><span>
Total time = 2t = 1.0261 sec</span>
