By calculating the crests, you can find the waves' frequency.
Hope this helps!
Answer:
The correct answer is Option A (decrease).
Explanation:
- According to Heisenberg's presumption of unpredictability, it's impossible to ascertain a quantum state viewpoint as well as momentum throughout tandem.
- Also, unless we have accurate estimations throughout the situation, we will have a decreased consistency throughout the velocity as well as vice versa though too.
Other given choices are not connected to the given query. Thus the above is the right answer.
The higher density allows the protogalactic clouds to cool faster and form an elliptical galaxy rather than a spiral galaxy.
In physical cosmology, a protogalaxy or protogalactic cloud , which could also be called a "primeval galaxy", is a cloud of gas which is forming into a galaxy. It is believed that the rate of star formation during this period of galactic evolution will determine whether a galaxy is a spiral or elliptical galaxy; a slower star formation tends to produce a spiral galaxy. The smaller clumps of gas in a protogalaxy form into stars.
Composition
Since there had been no previous star formation to create other elements, protogalaxies would have been made up almost entirely of hydrogen and helium. The hydrogen would bond to form H2 molecules, with some exceptions. This would change as star formation began and produced more elements through the process of nuclear fusion.
Mechanics
Once a protogalaxy begins to form, all particles bound by its gravity begin to free fall towards it. The time taken for this free-fall to conclude can be approximated using the free-fall equations. Most galaxies have completed this free-fall stage to become stable elliptical or disk galaxies, the disks taking longer to fully form. The formation of galaxy clusters takes much longer and is still in progress now.
This stage is also where galaxies acquire most of their angular momentum. A protogalaxy acquires this due to gravitational influence from neighbouring dense clumps in the early universe, and the further the gas is away from the centre, the more spin it gets.
Learn more about protogalactic clouds here : brainly.com/question/28166070
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Answer:
A.The spring constant for B is one quarter of the spring constant for A.
Explanation:
If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

As
, and the 2 springs have the same mass




So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m