Answer:
The length of the open closed pipe is 12.25 cm
Explanation:
In open - open organ pipe, third harmonic has Antinode to Node, Node to Node, Node to Node and Node to Antinode.
The length of the open-open organ pipe is equal to the sum of wavelength in Antinode to Node, Node to Node, Node to Node and Node to Antinode.
L = A→N + N→N + N→N + N→A
In open-closed pipe, Fundamental frequency has Antinode to Node.
Thus, length of the open-closed organ pipe is equal to the wavelength in Antinode to Node.
L = A→N
From the information in the question, fundamental frequency of open-closed pipe is to the third harmonic of the open-open pipe.
F₀ = F
Therefore, the length of the open closed pipe is 12.25 cm
Answer:
30.128 days
Explanation:
Given that:
Mean = 25
Standard deviation = 4
Confidence interval = 90% = 0.9
Since the confidence interval should not exceed 90%
Then using z test table
P(z) = 0.9
For 0.8997 , we get = 1.28
For 0.9015, we get = 1.29
∴
By solving
Z = 1.282
Thus, the duration to be used so that it will not exceed 90% C.I is:
Z = (x - μ)/σ
1.282= ( x - 25)/4
1.282 * 4 = x - 25
(1.282*4)+25 = x
x = 30.128 days
Answer:
a. Heat Capacity = 1.756J/mol-K
b. Heat Capacity = 24.942J/mol-k
Explanation:
Given
Constant volume Cv = 0.81J/mol-k
T1 = 34K
Td = Debye temperature = 306 K. Estimate the heat capacity (in J/mol-K) a. 44 K
First, The value of the temperature-independent constant.
Using Cv = AT³
Make A the subject of formula
A = Cv/T³
Substitute each values
A = 0.81/34³
A = 0.000020608589456543
A = 2.061 * 10^-5J/mol-k
The heat capacity changes with the temperature; below is the relationship between heat capacity and the temperature
Cv = AT³
So, The heat capacity when T = 44k is then calculated as
Cv = 2.061 * 10^-5 * 44³
Cv = 1.755522084266232
Cv = 1.756J/mol-K
(b) at 477 K.
Because the temperature is larger than the Debye temperature, the specific heat is calculated using as:
Cv = 3R
Where R = universal gas constant
R = 8.314J/mol-k
Cv = 3 * 8.314
Cv = 24.942J/mol-k
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