Answer:
The average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²
Explanation:
Given data,
The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec
= 3.1536 x 10⁷ +0.840
= 31536000.84 s
The period of 365 rotation of Earth in 2006, T₀ = 365 days
= 31536000 s
Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365
= 86400.0023 s
The time period of rotation is given by the formula,
Tₐ = 2π /ωₐ
ωₐ = 2π / Tₐ
Substituting the values,
ωₐ = 2π / 365.046306
= 7.272205023 x 10⁻⁵ rad/s
Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365
= 86400 s
Time period of rotation,
Tₓ = 2π /ωₓ
ωₓ = 2π / T
= 2π /86400
= 7.272205217 x 10⁻⁵ rad/s
The average angular acceleration
α = (ωₓ - ωₐ) / T₁
= (7.272205217 x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84
α = 6.152 X 10⁻²⁰ rad/s²
Hence the average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²
The average angular acceleration of the Earth from the year 1906 to 2006 would be -5.73 x 10^-20 rad/s^2. This value was obtained by finding the change in angular velocity and then dividing it by the elapsed time.
The question is asking for the average angular acceleration of the Earth from the year 1906 to 2006, during which the Earth's rotation rate decreased, causing the day to increase in duration by about 0.840 seconds.
To find the average angular acceleration, you first need to calculate the change in angular velocity, which can be found from the change in rotation time. One revolution (one day) is 2π radians, so the change in angular velocity is Δω = 2π/86400 s - 2π/(86400+0.840) s = -1.81 x 10^-10 rad/s.
The time interval from 1906 to 2006 is 100 years or about 3.16 x 10^9 seconds. Therefore, the average angular acceleration, α, which is the change in angular velocity divided by time, would be α = Δω/Δt = -1.81 x 10^-10 rad/s / 3.16 x 10^9 s = -5.73 x 10^-20 rad/s^2.
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the slinky particles move from the hand down the wave to the block
the slinky particles move side to side
the slinky particles move up and down
The characteristics of the traveling waves we can find the correct answer for the movement of the particles in the transverse wave is:
The traveling wave movement is a way of transmitting energy without mass movement, it is formed by two movements united, an oscillatory movement and a displacement movement, there are two possibilities:
In both cases, the matter particles are oscillating around their equilibriumposition and the wave is the one that has a displacement movement.
Let's review the claims:
a) False. The particle has an oscillatory motion, it does not have a net displacement
b) False. The movement is oscillatory
c) True. As the wave is transversal, the oscillatory movement from top to bottom.
In conclusion using the characteristics of the traveling waves we can find the correct answer for the motion of the particles in the transverse wave is:
Learn more here: brainly.com/question/14106293
Answer:
C.) The slinky particles move up and down
Explanation:
Transverse Wave-
A wave that has a disturbance perpendicular to the wave motion
Hello! This is the correct answer! Have a blessed day! :)
If you are in K12, please review the lesson! :) It will give you some very helpful definitions! I hope this helped!
B. protostar.
C. red giant.
D. nebula.
lens.?
Answer:
In a simple telescope two lenses are used, objective lens and eyepiece lens. A parallel beam of light is focused by the objective lens of a telescope and then light passes through an eyepiece forming a magnified image. Above diagram shows the ray diagram of an astronomical telescope.