<h2>Answer: The nucleus becomes entirely helium.</h2>
To shine, the stars transform their hydrogen into helium by means of <u>nuclear fusion</u>. When at half of its life a star is without hydrogen, the nucleus becomes entirely helium and <u>the star declines, becoming colder and brighter due to the energy generated by the nuclear reactions, then the star begins to contract.
</u>
Nevertheless, if the star is bigger, this helium will be also consumed and the nucleus transformed in Carbon, then in Oxigen, and so on. Being the last transformation Iron until the star delivers all its energy.
Note this process will depend on how massive the star is.
Answer:
Explanation:
This problem bothers on the energy stored in a spring in relation to conservation of energy
Given data
Mass of block m =200g
To kg= 200/1000= 0.2kg
Spring constant k = 1.4kN/m
=1400N/m
Compression x= 10cm
In meter x=10/100 = 0.1m
Using energy considerations or energy conservation principles
The potential energy stored in the spring equals the kinetic energy with which the block move away from the spring
Potential Energy stored in spring
P.E=1/2kx^2
Kinetic energy of the block
K.E =1/mv^2
Where v = velocity of the block
K.E=P.E (energy consideration)
1/2kx^2=1/mv^2
Kx^2= mv^2
Solving for v we have
v^2= (kx^2)/m
v^2= (1400*0.1^2)/0.2
v^2= (14)/0.2
v^2= 70
v= √70
v= 8.36m/s
a. Distance moved if the ramp exerts no force on the block
Is
S= v^2/2gsinθ
Assuming g= 9. 81m/s^2
S= (8.36)^2/2*9.81*sin60
S= 69.88/19.62*0.866
S= 69.88/16.99
S= 4.11m
Answer:
b) 4781 N
Explanation:
Because there is a redius do this question is talking about the acceleration force which= mv^2/r
so a=15^2/80=2.8125 m^2/s
so the force will be = m.a
F =1700×2.8125=4781.25 N
