Answer:
I.72m/s²
II.8m/s²
Explanation:
acceleration equal velocity² divided by length
Answer:
Explanation:
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion S = ut + 1/2at^2..... 1
For David u = 28.0 m/s where 'a' is set to nought
S = ut
S = 28t.......2
For Tina consider equation 1
Where acceleration = 2.90m/s^2 and u is set at nought
S = 1/2×2.90 m/s×t^2.......3
Equate 2 and 3
28t = 1.45t^2
Divide through by t
28 = 1.45t
t = 28/1.45
t = 19.31seconds
Now put the value of t into equation 3
S = 1/2×2.90 m/s×t^2.......3
= 1.45×20×20
= 580m
Tina must have driven 580meters before passing David
Considering the equation of linear motion : V^2 = U^2+2as
Where u is set at nought
V^2 = 2as
V^2 = 2×2.9×580
V^2 = 3364
V = √3364
V = 58m/s
Her speed will be 58m/s
Answer:
pascal
Explanation:
its obtained after either division or multiplication
Let's assume that ground level is the height 0 meters. The change in potential energy is going to be gravitational potential energy, which is given by PE=mgh.
ΔPE=mgh-mgy
=mg(h-y)
=50(28-0)
=1400 J
In stars more massive than the sun, the core temperature is hotter, which allows for fusion of more complex elements.
Most of the fusion occurs in the core.
In stars more massive than the sun, fusion continues through Deuterium, Carbon, and finally reaching iron/nickel.
Up to this point, the fusion reaction was endothermic, which means that the energy expended to produce the fusion reaction was exceeded by the energy produced in the reaction.
Fusion past iron is exothermic, and therefore the star will be able to survive by fusing elements heavier than iron.
After the core is almost entirely iron, the star is no longer in the Main Sequence.
So, fusion in stars more massive than the sun continue fusing until the core is almost entirely <em>iron</em>.