Structural constraint is the answer :)
Answer:
It's mostly known that time stops moving in a black hole, as for space, its known the spacetime changes over time. A black hole in such a state is essentially stationary. So for my research, time does not stand still in space unless were taking about black holes.
Explanation:
The only graph that accurately depict the given motion is graph D.
The given parameters;
- initial position of the man = 0
- direction of the man's first displacement = backward
- time of first motion, t₁ = 6 seconds
- velocity of this first displacement = v₁
- time without any motion (<em>zero movement</em>) = 6 seconds
- direction of the second displacement = forward
- velocity of second displacement = 2v₁
Let the acceleration of the first displacement = a
Acceleration of the second displacement = 2a
From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.
The only options with initial motion towards the negative direction are;
The difference between graph B and D;
- in graph B there is a uniform motion for 6 seconds
- in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).
Thus, the only graph that accurately depict the given motion is graph D.
Learn more here: brainly.com/question/21095906
Answer:
They both describe atoms as being made up of positive and negative matter.
Explanation:
In both Bohr's model and Thomson model, the atom consists of positively-charged matter and negatively-charged matter. However, the structure of the atom in the two models is totally different:
- in Thomson's model, the atom consists of a large sphere of uniform positive charge, and electrons (which are negatively charged) are scattered all around inside this sphere
- In Bohr's model, the atom consists of a small, positively charged nucleus, while the electrons (negatively charged) orbit around the nucleus in precise orbits.
Solution :
Given :
Wavelength of the thin beam of light, λ = 50 μm
Distance of the screen from the slit, D = 3.00 m
Width of the fringe, Δy = ±8.24 mm
Therefore, width of the slit is given by :
= 0.000018203 m
= 0.0182 mm
= 0.018 mm
The intensity of light is given by :
, where 
Now, 
= 0.1854
≈ 0.18
= 2 x0.81
