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Mashutka [201]
3 years ago
12

A strip of copper metal is riveted to a strip of aluminum. the two metals are then heated. the coefficient of linear expansion o

f aluminum is greater than that of copper. what happens?
Physics
1 answer:
olganol [36]3 years ago
4 0
We can base this on the equation of thermal expansion.

ΔL = L₀αΔT
where
ΔL is the expansion of length upon heating
L₀ is the initial length
α is the coefficient of linear expansion
ΔT is the temperature difference

So, if α for Aluminum is greater than α for Copper, then after heating, aluminum would be longer than copper.
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Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea
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The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Where,

L = Vibrating length string

T = Tension in the string

\mu = Linear mass density

At the same time we have the expression for the number of beats described as

n = |f_1-f_2|

Where

f_1 = First frequency

f_2 = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

f \propto \sqrt{T}

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

n = f_2-f_1

f_2 = n+f_1

Replacing 3/sfor n and 202Hz for f_1,

f_2 = 3/s + 202Hz

f_2 = 3/s(\frac{1Hz}{1/s})+202Hz

f_2 = 206Hz

The frequency of the tightened is 205Hz

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