When you drop a 0.4 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s° toward the earth's surface. According to Newton's third law, the apple must exert an equal but opposite force on Earth.If the mass of the earth 5.98 x 1024 kg, what is the magnitude of the earth's acceleration toward the apple?
Answer in units of m/s?.
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If the mass of the earth 5.98 x 1024 kg, what is the magnitude of the earth's acceleration toward the apple?
Answer in units of m/s?.
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Answer:
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Explanation:
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Answer : There exist an inverse relationship between the wavelength and the energy of the wave.
Explanation :
Ultraviolet radiation is dangerous because it has a high enough energy to damage skin cells.
The energy of a wave is given by the following relation :
So, it is clear that relation between the wavelength and the energy of the wave is inverse.
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We will calculate the percentage of uncertainties.
For a length: p 1 = ( 0.2 · 100 ) : 22.4 = 20 : 22.4 = 0.893
For a width: p 2 = ( 0.1 · 100 ) : 8.4 = 10 : 8.4 = 1.190
p 1 + p 2 = 0.893 + 1.190 = 2.083 %
A = 22.4 · 8.4 = 188.16 +/- 2.083 %
Uncertainty: 188.16 · 0.02083 = 3.92
A = 188.16 +/- 3.92 cm²
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Explanation:
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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
Final answer:
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.
Explanation:
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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