Answer;
D. where two plates collide
Explanation;
-Subduction zones are plate tectonic boundaries where two plates converge, and one plate is thrust beneath the other. This process results in geohazards, such as earthquakes and volcanoes.
-Subduction zone volcanism occurs where two plates are converging on one another. One plate containing oceanic lithosphere descends beneath the adjacent plate, thus consuming the oceanic lithosphere into the earth's mantle. This on-going process is called subduction.
The kinetic energy is the same as the potential energy of raising it 40cm (0.4m). That's mgh where m is mass of ball. Its then 3.924*m, whatever m is equal to in kg.
Venus is called Earth's "Twin" because Earth and Venus have almost the same mass, size, similar composition and they are neighboring planets. Some of most notable differences though between the two planets are their color, temperature, Earth's ability to support life and their atmosphere. Venus' atmosphere is about 100x thicker than Earth's, Earth can support life while Venus can not because Earth has water and plant life and Earth is a bright blue and green while Venus is more orange and red.
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.
Answer:
There are <u>5</u> significant figures.
Explanation:
You must start conting your sig figs until you continue to hit zeros at the end. Those zeroes at the end are disregarded. So 0.0609 is where you get your <em>sig figs</em> from.
