In the picture at left. The launch height is 10m ball B mass 5kg has a launch velocity of 12 m/s and ball A mass 8 kg has a laun
1 answer:
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Answer:
(a)
(b)
Explanation:
Given data
The source is 1.00 μW
The point is 3m away from the point source
For Part (a)
The intensity at a certain distance from a point source that emits sound wave is given as:
For Part (b)
The Sound level is given by
β=(10dB)×log(I/I₀)
Where
I₀=1×10⁻¹²W/m²
Substitute the given values to find Sound level
So
Answer:
<em>The hotter star radiates 625 times more energy per second from each square meter of its surface</em>
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Explanation:
Temperature of the hotter star is 30000 K
temperature of the cooler star = 6000 K
From Stefan-Boltzmann radiation laws, for a non black body
P = εσA
where
P is the energy per second or power of radiation
ε is the emissivity of the body
σ is the Stefan-Boltzmann constant of proportionality
A is the area of the sun
T is the temperature of the sun
The sun can be approximated as a black body, and the equation reduces to
P = σA
For the hotter body,
P = σA() = 8.1 x 10^17σA J/s
For the cooler body,
P = σA() = 1.296 x 10^15σA J/s
comparing the two stars energy
==> (8.1 x 10^17)/(1.296 x 10^15) = 625
<em>This means that the hotter star radiates 625 times more energy per second from each square meter of its surface</em>
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