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3 years ago

12

What amount of heat is removed to lower the temperature of 80 grams of water from 75ºC to 45ºC? The specific heat of liquid wate

r is 4.18 J/gºC.

2 answers:

chubhunter [2.5K]3 years ago

5 0

=4.18*80*(75-45)

=4.18*80*30

=10032

amm18123 years ago

3 0

amount of heat is removed to lower the temperature of 80 grams of water from 75ºC to 45ºC with specific heat of liquid water is 4.18 J/gºC

= 4.18 J/gºC * 80g * (75-45)ºC

= 10032J

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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

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(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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A house is maintained at 1 atm and 24°C, and warm air inside a house is forced to leave the house at a rate of 90 m3/h as a resu

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A student used a tuning fork of frequency 320 Hz and observed that the speed of sound was 339 m/s. Calculate the wavelength of t

Helen [10]

To solve this problem it is necessary to apply the concepts related to wavelength as a function of speed and frequency. In mathematical terms it can be expressed as

$\lambda = \frac{v}{f}$

Where,

v = Velocity

f = Frequency

According to our values the frequency (f) is 320Hz and the speed (v) is 339m / s.

Replacing in the given equation we have to,

$\lambda = \frac{v}{f}\\\lambda = \frac{339}{320}\\\lambda = 1.059m\approx 1.06m$

Therefore the wavelength of this sound wave is 1.06m

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3 years ago

If 5 complete oscillations of a sound wave pass through a point in 0.5 s and the speed of sound was recorded to be 10 m/s, then

LekaFEV [45]

Answer:

λ = 2.5m

Explanation:

Given the following :

Speed of sound (v) = 10m/s

If 5 oscillations pass through a point in 0.5seconds;

Time taken (period) for 1 oscillation is :

Number of oscillations / total time taken

5 / 0.5 = 0.25 seconds

Wavelength, period and Velocity are related by the formula:

v = λ / T

λ = v * T

λ = 10 * 0.25

λ = 2.5 m

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3 years ago