Answer:
gₓ = 6.52 m/s²
Explanation:
The value of acceleration due to gravity on the surface of earth is given as:
g = GM/R² -------------------- equation 1
where,
g = acceleration due to gravity on surface of earth
G = Universal Gravitational Constant
M = Mass of Earth
R = Radius of Earth
Now, for the alien planet:
gₓ = GMₓ/Rₓ²
where,
gₓ = acceleration due to gravity at the surface of alien planet
Mₓ = Mass of Alien Planet = 2.4 M
Rₓ = Radius of Alien Planet = 1.9 R
Therefore,
gₓ = G(2.4 M)/(1.9 R)²
gₓ = 0.66 GM/R²
using equation 1
gₓ = 0.66 g
gₓ = (0.66)(9.81 m/s²)
<u>gₓ = 6.52 m/s²</u>
The angular frequency of this motion is 5.46 rad/s.
The oscillation of spring is an example of Simple Harmonic Motion(SHM).
The general equation of an SHM is given by the formula.
X = Acos(wt)
Here A is the amplitude
ω is the angular frequency
T is the time
Comparing the above equation with the given condition,
X = 17.4 cm cos(5.46t)
A = 17.4 cm
ω = 5.46 rad/s
T = 1 s
Hence, the angular frequency of this motion is 5.46 rad/s.
To know more about the "general equation of SHM", refer to the link below:
brainly.com/question/14869852?referrer=searchResults
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d = 400 m (8 laps) = 2400 m
t = 14.5 min
v = 0
average t = 14.5 min (60 sec per min) = 870 s/8 laps = 108.75 s per lap
speed = distance /time
400/108.75 sec = 3.67 m/s
Answer:
r₂ = 1,586 m
Explanation:
For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m
β = 10 log (I / I₀)
where Io is the sensitivity threshold 10⁻¹² W / m²
I₁ / I₀ =
I₁ = I₀ e^{\beta/10}
let's calculate
I₁ = 10⁻¹² e^{25/10}
I₁ = 1.20 10⁻¹¹ W / m²
the other intensity in exercise is
I₂ = 10⁻¹² e^{80/10}
I₂ = 2.98 10⁻⁹ W / m²
now we use the definition of sound intensity
I = P / A
where P is the emitted power that is a constant and A the area of the sphere where the sound is distributed
P = I A
the area a sphere is
A = 4π r²
we can write this equation for two points of the found intensities
I₁ A₁ = I₂ A₂
where index 1 corresponds to 25m and index 2 to the other distance
I₁ 4π r₁² = I₂ 4π r₂²
I₁ r₁² = I₂ r₂²
r₂ = √ (I₁ / I₂) r₁
let's calculate
r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25
r₂ = √ (0.40268 10⁻²) 25
r₂ = 1,586 m