The answer is 125 Joules
The first thing to take note of is the work equation: W=F×D
Since we already have our force and our distance that will help make this problem easier.
So, W=25*5
W=125
Therefore, our answer is 125 Joules since work is measured in joules
Hope this helped!! :)
Venus has a dense atmosphere of mostly carbon dioxide. <em>(D)</em>
A, B, and C are false statements.
Answer:
Option D
670 Kg.m/s
Explanation:
Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)
Final momentum is also mv but v being westward direction, we take it negative
Final momentum=82*-2.5= -205 Kg.m/s
Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s
Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s
Answer:
mass is in kg and velocity is in m/s for the answer to be in Joules
Explanation:
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.
