Answer:
799.54 ft
Explanation:
Linear thermal expansion is:
ΔL = α L₀ ΔT
where ΔL is the change in length,
α is the linear thermal expansion coefficient,
L₀ is the original length,
and ΔT is the change in temperature.
Given:
α = 1.2×10⁻⁵ / °C
L₀ = 800 ft
ΔT = -17°C − 31°C = -48°C
Find: ΔL
ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)
ΔL = -0.4608
Rounded to two significant figures, the change in length is -0.46 ft.
Therefore, the final length is approximately 800 ft − 0.46 ft = 799.54 ft.
Generated for what there no waves shown
Given :
Object A is 71 degrees and object B is 75 degrees .
To Find :
How will thermal energy flow.
Solution :
We know, by law of thermodynamics thermal energy will flow from higher temperature to lower temperature.
So, in the given question energy will flow from object B from object A.
Hence, this is the required solution.
is the intensity of the sound.
Answer: Option B
<u>Explanation:</u>
The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about
). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.
This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

Where,
I = Intensity of the sound produced
= Standard Intensity of sound of 60 decibels = 
So for 19 decibels, determine I as follows,



When log goes to other side, express in 10 to the power of that side value,


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