The amount of diffraction depends on the wavelength of light, with shorter wavelengths being diffracted at a greater angle than longer ones (in effect, blue and violet<span> light are diffracted at a larger angle than is red light).
I hope my answer has come to your help. God bless and have a nice day ahead!
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Answer:
When did humans learn that the Earth is not the center of the universe?
Answer
1
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4 Answers
Asked in 3 Spaces


Science - Next Generation
Alexander Somm
, Consultant, Investor Relations at Novelpharm AG (2015-present)
Answered Oct 16
What, it isn’t?!
Sorry, I had to.
As far as I have read and understood, the Sumerians and later the Babylonians both had astronomical calendars that already differentiated planets and stars. Earth was not the center to them, the Sun likely was. That was around 2,200 - 1,600 BC.
After that, Greek philosopher Aristarchus of Samos (310 - 230 BC) was the first (recorded) to have believed the solar system was organized around the Sun, rather than the Earth. His heliocentric model was unpopular during Aristarchus’ lifetime, although it would inspire astronomers centuries later, such as Copernicus and Galileo.
Now, there are numerous archeological findings (cave paintings) and studies, that all suggest an understanding of complex astronomy in prehistoric times dating back as far as 40,000 years. This also explains how early, prehistoric migrants may have navigated the seas.
Explanation:
hope it helps
have a good day
Answer:
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Explanation:
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Answer: Volume = 1080m^3
Explanation:
Given that the prism has a 15 m by 18 m rectangular base and a height of 4 m
Volume is the product of length, breath and height. That is
Volume = L × B × H
Where
L = 18 m
B = 15m
H = 4m
Using the formula above gives:
Volume V = 18 × 15 × 4
V = 1080 m^3
Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
