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.
Let's cut through the weeds and the trash
and get down to the real situation:
A stone is tossed straight up at 5.89 m/s .
Ignore air resistance.
Gravity slows down the speed of any rising object by 9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for
(5.89 m/s / 9.8 m/s²) = 0.6 seconds.
At that timer, he has run out of upward gas. He is at the top
of his rise, he stops rising, and begins to fall.
His average speed on the way up is (1/2) (5.89 + 0) = 2.945 m/s .
Moving for 0.6 seconds at an average speed of 2.945 m/s,
he topped out at
(2.945 m/s) (0.6 s) = 1.767 meters above the trampoline.
With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.
(0.6 sec up) + (0.6 sec down) = 1.2 seconds until he hits rubber again.
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
Here mass of the iron pan is given as 1 kg
now let say its specific heat capacity is given as "s"
also its temperature rise is given from 20 degree C to 250 degree C
so heat required to change its temperature will be given as
now if we give same amount of heat to another pan of greater specific heat
so let say the specific heat of another pan is s'
now the increase in temperature of another pan will be given as
now we have
now as we know that s' is more than s so the ratio of s and s' will be less than 1
And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature
So correct answer is
<u>A) The second pan would reach a lower temperature.</u>
The work done by the applied force on the block against the frictional force is 15.75 J.
<h3>
Work done by the applied force</h3>
The work done by the applied force is calculated as follows;
W = Fd
F - Ff = ma
where;
- F is applied force
- Ff is frictional force
Fcos(37) - μmgsin(37) = ma
Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)
0.799F - 7.077 = 0.8
F = 9.86 N
W = Fdcosθ
W = 9.86 x 2 x cos(37)
W = 15.75 J
Thus, the work done by the applied force on the block against the frictional force is 15.75 J.
Learn more about work done here: brainly.com/question/25573309
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