As the core collapses, the outer layers of the star are expelled. A planetary nebula is formed by the outer layers. The core remains as a white dwarf and eventually cools to become a black dwarf. ... Like low-mass stars, high-mass stars are born in nebulae and evolve and live in the Main Sequence
hydrogen shell burning - outer layers swell. Red Giant Branch - helium ash core compresses - increased hydrogen shell burning. First Dredge Up - expanding atmosphere cools star - stirs carbon, nitrogen and oxygen upward - star heats up.
As close as I can read it, it appears to be
1/12 gram/second
(0.08333... gm/sec)
Answer:
2.32 s
Explanation:
Using the equation of motion,
s = ut+g't²/2............................ Equation 1
Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.
Note: Since Onur drops the basket ball from a height, u = 0 m/s
Then,
s = g't²/2
make t the subject of the equation,
t = √(2s/g')...................... Equation 2
Given: s = 10 m, g' = 3.7 m/s²
Substitute this value into equation 2
t = √(2×10/3.7)
t = √(20/3.7)
t = √(5.405)
t = 2.32 s.
Answer:
1.71 km
Explanation:
Convert 30 minutes to seconds:
30 min × (60 s / min) = 1800 s
Find the displacement:
0.95 m/s × 1800 s = 1710 m
Convert to kilometers:
1710 m × (1 km / 1000 m) = 1.71 km
Answer:
3.28 m
3.28 s
Explanation:
We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.
Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = 0
V0 = 4 m/s
a = -2.45 m/s^2 (because the acceleration is down slope)
Then:
X(t) = 4*t - 1.22*t^2
And the equation for speed is:
V(t) = V0 + a * t
V(t) = 4 - 2.45 * t
If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:
0 = 4 - 2.45 * t
4 = 2.45 * t
t = 1.63 s
Replacing that time on the position equation:
X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m
To find the time it will take to return we equate the position equation to zero:
0 = 4 * t - 1.22 * t^2
Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:
0 = t * (4 - 1.22*t)
t1 = 0
0 = 4 - 1.22*t2
1.22 * t2 = 4
t2 = 3.28 s
