Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.
Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m
Now you need the final speed to use it as initial speed of the next part.
Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s
Part B) Free fall
Maximum height, y max ==> Vf = 0
Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s
ymax = yo + Vo*t - g[t^2] / 2
ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m
Answer: ymax = 10084.2m
Answer:
λ = 162 10⁻⁷ m
Explanation:
Bohr's model for the hydrogen atom gives energy by the equation
= - k²e² / 2m (1 / n²)
Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer
The Planck equation
E = h f
The speed of light is
c = λ f
E = h c /λ
For a transition between two states we have
-
= - k²e² / 2m (1 /
² -1 /
²)
h c / λ = -k² e² / 2m (1 /
² - 1/
²)
1 / λ = (- k² e² / 2m h c) (1 /
² - 1/
²)
The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses
Let's calculate the emission of the transition
1 /λ = 1.097 10⁷ (1/10² - 1/8²)
1 / λ = 1.097 10⁷ (0.01 - 0.015625)
1 /λ = 0.006170625 10⁷
λ = 162 10⁻⁷ m
Answer:
C sedimentary rocks is the answer.
Explanation:
Answer: an isolated system
Explanation: