Electrons are transferred sequentially between the two photosystems, with photosystem I acting to generate NADPH and photosystem II acting to generate ATP. The pathway of electron flow starts at photosystem II, which is homologous to the photosynthetic reaction center of R. viridis already described.
Multiply the masses by the respective distances:
(12 kg) (2 m) = 24 J
(25 kg) (1 m) = 25 J
so the heavier bag takes more work to lift, and (b) is the answer.
(d) is technically correct if the sacks are carrying different contents whose masses are not equal, but since we don't know what's inside each sack, assume 12 kg and 25 kg are the masses of each sack *and* their contents.
I was going to beg off until tomorrow, but this one is nothing like those others.
Why, at only 40km/hr, we can ignore any relativistic correction, and just go with Newton.
To put a finer point on it, let's give the car a direction. Say it's driving North.
a). From the point of view of the car, its driver, and passengers if any,
the pole moves past them, heading south, at 40 km/hour .
b). From the point of view of the pole, and any bugs or birds that may be
sitting on it at the moment, the car and its contents whiz past them, heading
north, at 40 km/hour.
c). A train, steaming North at 80 km/hour on a track that exactly parallels
the road, overtakes and passes the car at just about the same time as
the drama in (a) and (b) above is unfolding.
The rail motorman, fireman, and conductor all agree on what they have
seen. From their point of view, they see the car moving south at 40 km/hr,
and the pole moving south at 80 km/hr.
Now follow me here . . .
The car and the pole are both seen to be moving south. BUT ... Since the
pole is moving south faster than the car is, it easily overtakes the car, and
passes it . . . going south.
That's what everybody on the train sees.
==============================================
Finally ... since you posed this question as having something to do with your
fixation on Relativity, there's one more question that needs to be considered
before we can put this whole thing away:
You glibly stated in the question that the car is driving along at 40 km/hour ...
AS IF we didn't need to know with respect to what, or in whose reference frame.
Now I ask you ... was that sloppy or what ? ! ?
Of course, I came along later and did the same thing with the train, but I am
not here to make fun of myself ! Only of others.
The point is . . . the whole purpose of this question, obviously, is to get the student accustomed to the concept that speed has no meaning in and of itself, only relative to something else. And if the given speed of the car ...40 km/hour ... was measured relative to anything else but the ground on which it drove, as we assumed it was, then all of the answers in (a) and (b) could have been different.
And now I believe that I have adequately milked this one for 50 points worth.
Answer:
option C
Explanation:
Final velocity of the object is 114 m / s. Hence, final velocity of the object is 114 m / s.
Answer:
a= 3.49 m/s^2
Explanation:
magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.
we know that tangential acceleration a_t= change in velocity /time taken
now 90 km/h = 25 m/s
a_t = 25/17 = 1.47 m/s^2.
radial acceleration a_r = v^2/r
v= a_t×t = 1.47×13 = 19.11 m/s
a_r = 19.11^2/115= 3.175
now,
a= 3.49 m/s^2
