Answer:
Ek1 = 900000 [J]
Ek1 = 400000 [J]
Explanation:
In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

where:
m = mass = 500 [kg]
v1 = 60 [m/s]
So we have:
Ek1 = 0.5*500*(60^2)
Ek1 = 900000 [J]
and:
Ek2 = 0.5*500*(40^2)
Ek2 = 400000 [J]
Energy can neither be created nor be destroyed.
Answer:
Protons and Neutrons are found in the nucleus and the electrons are found on the outside of the nucleus
Explanation:
Hope this helps
Answer:
600,000,000 degree C
Explanation:
This stage is the last stage and is refereed to as supernova. In the beginning of this stage, gravity pulls the inner core and crush it, due to which fusion of atoms starts. Carbon and Oxygen fuse together and the temperature is about of 600,000,000 degree C.
The most heavier atom that can be formed out of this fusion is the iron. The moment all the atoms becomes of iron, no further fusion is possible hence that body emits radiation of high intensity and collapse causing a big supernova.
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s