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dexar [7]
4 years ago
8

The process of generating electricity from dams and nuclear power plants is very different, but some of the energy transformatio

ns are the same. which energy transformations occur in both dams and nuclear power plants? check all that apply. kinetic energy to mechanical energy nuclear energy to thermal energy mechanical energy to electrical energy thermal energy to electrical energy
Physics
2 answers:
storchak [24]4 years ago
8 0
There are only two correct options. Kinetic energy to mechanical energy, and mechanical energy to electrical energy. 
julia-pushkina [17]4 years ago
6 0

Answer:

<em>Kinetic energy to mechanical energy </em> and <em>mechanical to electrical energy </em>

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4 years ago
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Two strings on a musical instrument are tuned to play at 196 hz (g) and 523 hz (c). (a) what are the first two overtones for eac
Tems11 [23]
(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
f_n = n f_1
where f1 is the fundamental frequency.

So, the first overtone (2nd harmonic) of the string is
f_2 = 2 f_1 = 2 \cdot 196 Hz = 392 Hz
while the second overtone (3rd harmonic) is
f_3 = 3 f_1 = 3 \cdot 196 Hz = 588 Hz

Similarly, for the second string with fundamental frequency f_1 = 523 Hz, the first overtone is
f_2 = 2 f_1 = 2 \cdot 523 Hz = 1046 Hz
and the second overtone is
f_3 = 3 f_1 = 3 \cdot 523 Hz = 1569 Hz

(b) The fundamental frequency of a string is given by
f=  \frac{1}{2L}  \sqrt{ \frac{T}{\mu} }
where L is the string length, T the tension, and \mu = m/L is the mass per unit of length. This part  of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them m_{196}, the mass of the string of frequency 196 Hz, and m_{523}, the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
\frac{523 Hz}{196 Hz} =  \frac{ \frac{1}{2L}  \sqrt{ \frac{T}{m_{523}/L} } }{\frac{1}{2L}  \sqrt{ \frac{T}{m_{196}/L} }}
and since L and T simplify in the equation, we can find the ratio between the two masses:
\frac{m_{196}}{m_{523}}=( \frac{523 Hz}{196 Hz} )^2 = 7.1

(c) Now the tension T and the mass per unit of length \mu is the same for the strings, while the lengths are different (let's call them L_{196} and L_{523}). Let's write again the ratio between the two fundamental frequencies
\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L_{523}} \sqrt{ \frac{T}{\mu} } }{\frac{1}{2L_{196}} \sqrt{ \frac{T}{\mu} }} 
And since T and \mu simplify, we get the ratio between the two lengths:
\frac{L_{196}}{L_{523}}= \frac{523 Hz}{196 Hz}=2.67

(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them T_{196} and T_{523}. Let's write again the ratio of the frequencies:
\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L} \sqrt{ \frac{T_{523}}{m/L} } }{\frac{1}{2L} \sqrt{ \frac{T_{196}}{m/L} }}
Now m and L simplify, and we get the ratio between the two tensions:
\frac{T_{196}}{T_{523}}=( \frac{196 Hz}{523 Hz} )^2=0.14
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