An object of mass m is oscillating back and forth in simple harmonic motion. The maximum origin, x = 0, and moving in the x dire
1 answer:
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Answer:
4.7 s
Explanation:
The complete question is presented in the attached image to this solution.
v(t) = 61 - 61e⁻⁰•²⁶ᵗ
At what time will v(t) = 43 m/s?
We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.
43 = 61 - 61e⁻⁰•²⁶ᵗ
- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18
e⁻⁰•²⁶ᵗ = (18/61) = 0.2951
In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205
-0.26t = -1.2205
t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.
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Answer:
The speed of the resistive force is 42.426 m/s
Explanation:
Given;
mass of skydiver, m = 75 kg
terminal velocity,
The resistive force on the skydiver is known as drag force.
Drag force is directly proportional to square of terminal velocity.
Where;
k is a constant
When the new drag force is half of the original drag force;
Therefore, the speed of the resistive force is 42.426 m/s
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