A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen
1 answer:
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Answer:
option E
Explanation:
given,
diameter = 4 mm
shutter speed = 1/1000 s
diameter of aperture = ?
shutter speed = 1/250 s
exposure time to the shutter time
N is the diameter of the aperture and t is the time of exposure
now,
inserting all the values
N₂² = 4
N₂ = 2 mm
hence , the correct answer is option E
Answer:
at the speed of light ()
Explanation:
The second postulate of the theory of the special relativity from Einstein states that:
"The speed of light in free space has the same value c in all inertial frames of reference, where "
This means that it doesn't matter if the observer is moving or not relative to the source of ligth: he will always observe light moving at the same speed, c.
In this problem, we have a starship emitting a laser beam (which is an electromagnetic wave, so it travels at the speed of light). The startship is moving relative to the Earth with a speed of 2.0*10^8 m/s: however, this is irrelevant for the exercise, because according to the postulate we mentioned above, an observer on Earth will observe the laser beam approaching Earth with a speed of .
Answer:
(i) 12 seconds
(ii) 216 meters from the initial position
(iii) 132 meters from the initial position
(iv) No
Explanation:
Speed of express train =36 m/s
Speed of local train =11 m/s
The initial distance between the local train and passenger train =100 m.
Due to the application of breaks, the express train slows at the rare of
So, the acceleration of the express train, .
(i) Let t be the time the express train takes to stop.
From the equation of motion,
v=u+at
where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.
In this case, v=0, u=36 m/s,
So, 0=36+(-3)t
seconds.
(ii) To compute the distance traveled, s, till the express train stops, using
meters.
(iii) The local train is moving at a speed of 11 m/s
So, in 12 seconds, the distance, d, traveled by the local train
d= 11x12=132 meters [as distance= speed x time]
(iv) Let 0 be the reference position which is the initial position of the express train.
So, at the initial time, the position of the local train is at 100m.
After 12 seconds:
The position of the express train is at 216 m [using part (ii)]
and the position of the local train is at 100+132=232m [using part (iii)].
So, the local train is still ahead of the express train, hence the trains didn't collide.