Answer:
0.11m
Explanation:
let's assume the boat is of uniform construction
Ignoring friction losses
Also assume the origin is at the end of the boat originally with the heavier person
the center of mass of the whole system will not change relative to the water when the two swap ends
Originally, the center of mass is
85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin
after the swap, the center of mass is
50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin
The center of mass has shifted
1.14-1.030 = 0.11m
as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat
<span>A) x = 41t
The classic equation for distance is velocity multiplied by time. And unfortunately, all of your available options have the form of that equation. In fact, the only difference between any of the equations is what looks to be velocity. And in order to solve the problem initially, you need to divide the velocity vector into a vertical velocity vector and a horizontal velocity vector. And the horizontal velocity vector is simply the cosine of the angle multiplied by the total velocity. So
H = 120*cos(70) = 120*0.34202 = 41.04242
So the horizontal velocity is about 41 m/s. Looking at the available options, only "A" even comes close.</span>
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:
d= 0.242 mm
Explanation:
Slit width (d ) = ?
Screen distance ( D ) = 1.25 m
Wave length of light λ = 600 nm
Distance of n the dark fringe from centre
= n λ D / d
Here n = 2
so
d= 0.242 mm
Answer:
Use the ammeter to measure the current that flows through each wire, because a larger current that flows through the wire corresponds to a smaller resistivity
Explanation:
Since they are connected to a constant voltage power source, the potential difference does not change. The potential difference is proportional to the product of the current and the resistance and, the resistance opposes the flow of electric current. It is clear to see that a large current that flows through the current means there is a lesser resistance to the flow of current at constant potential difference across the circuit.
