Answer:
5.38 m/s
Explanation:
Given (in the x direction):
Δx = 2.45 m
v₀ = v cos 42.5°
a = 0 m/s²
Δx = v₀ t + ½ at²
(2.45 m) = (v cos 42.5°) t + ½ (0 m/s²) t²
2.45 = (v cos 42.5°) t
t = 3.32 / v
Given (in the y direction):
Δy = 0.373 m
v₀ = v sin 42.5°
a = -9.8 m/s²
Δx = v₀ t + ½ at²
(0.373 m) = (v sin 42.5°) t + ½ (-9.81 m/s²) t²
0.373 = (v sin 42.5°) t − 4.905 t²
0.373 = (v sin 42.5°) (3.32 / v) − 4.905 (3.32 / v)²
0.373 = 2.25 − 54.2 / v²
v = 5.38
Graph:
desmos.com/calculator/5n30oxqmuu
Gas Clouds (of Hydrogen), starts contracting under the influence of gravitational pull of gas.
As it's contracting, it's pressure & temperature increases, it is called "Protostar" now.
When Temperature become sufficiently high (4 million C), Nuclear fusion starts, & large amount of energy releases.
Energy Travels to star's surface and is radiated in the form of light, heat & EM Radiation.
The contraction of star stops only when "Inward Gravitational pull" is balanced by "Outward Radiant Energy". At this time, star becomes stable in size & temperature.
Contraction continues, 'causes star's outer region to 'boil' & 'expand'. At this stage, it becomes brightest, red giant.
As fuel burns, helium nuclei fuses & form Carbon nuclei further to silicon & further to Iron Nuclei by progressive Nuclear Fusions. At all this stages, star was very unstable.
Star end it's life in many different way & it depends on it's mass, whether it ends as "White Dwarf", "Neutron Star" or "Supernova"
This was a short note on life story of star. <span>Hope this helps!</span>
Answer:
vo = 0.175m/s
a = -0.040625 m/s^2
Explanation:
To solve this problem, you will need to use the equations for constant acceleration motion:
In the first equation you relate final position with the time elapsed, in the second one, you relate final velocity at any given position. In both equations, you will have both the acceleration a and the initial velocity vo as variables. We can simplify with the information we have:
1. 
2. 
Replacing in the first equation:
But as you are told that the ball was projected om the air track, it only makes sense for the velocity to be positive, otherwise it would have started moving outside the air track, so the real solution is 0.175m/s. Then, the acceleration would be:
Explanation:
We need to find the speed of a 3.1eV proton and a 3.1eV electron.
For proton, using conservation of energy such that,
is mass of proton
For electron,
is mass of proton
Hence, this is the required solution.
