Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae
------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e![^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}](https://tex.z-dn.net/?f=%5E%7B%5B%28-93.1%2A10%5E3%29%28J%2Fmol%29%5D%2F%5B%288.314%29%28J%2Fmol.K%29%28332K%29%7D)
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))
Given:
![X_{L} = 50.0 \ohm](https://tex.z-dn.net/?f=X_%7BL%7D%20%3D%2050.0%20%5Cohm)
frequency, f = 60.0 Hz
frequency, f' = 45.0 Hz
![V_rms} = 85.0 V](https://tex.z-dn.net/?f=V_rms%7D%20%3D%2085.0%20V)
Solution:
To calculate max current in inductor,
:
At f = 60.0 Hz
![X_{L} = 2\pi fL](https://tex.z-dn.net/?f=X_%7BL%7D%20%3D%202%5Cpi%20fL)
![50.0 = 2\pi\times 60.0\times L](https://tex.z-dn.net/?f=50.0%20%3D%202%5Cpi%5Ctimes%2060.0%5Ctimes%20L)
L = 0.1326 H
Now, reactance
at f' = 45.0 Hz:
![X'_{L} = 2\pi f'L](https://tex.z-dn.net/?f=X%27_%7BL%7D%20%3D%202%5Cpi%20f%27L)
![X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm](https://tex.z-dn.net/?f=X%27_%7BL%7D%20%3D%202%5Cpi%5Ctimes%2045.0%5Ctimes%200.13263%20%3D%2037.5%5Cohm%20)
Now,
is given by:
Therefore, max current in the inductor,
= 2.13 A
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
![D_o=11.9inch](https://tex.z-dn.net/?f=D_o%3D11.9inch)
Explanation:
From the question we are told that:
Thickness ![T=0.5](https://tex.z-dn.net/?f=T%3D0.5)
Internal Pressure![P=2.2Ksi](https://tex.z-dn.net/?f=P%3D2.2Ksi)
Shear stress ![\sigma=12ksi](https://tex.z-dn.net/?f=%5Csigma%3D12ksi)
Elastic modulus ![\gamma= 35000](https://tex.z-dn.net/?f=%5Cgamma%3D%2035000)
Generally the equation for shear stress is mathematically given by
![\sigma=\frac{P*r_1}{2*t}](https://tex.z-dn.net/?f=%5Csigma%3D%5Cfrac%7BP%2Ar_1%7D%7B2%2At%7D)
Where
r_i=internal Radius
Therefore
![12=\frac{2.2*r_1}{2*0.5}](https://tex.z-dn.net/?f=12%3D%5Cfrac%7B2.2%2Ar_1%7D%7B2%2A0.5%7D)
![r_i=5.45](https://tex.z-dn.net/?f=r_i%3D5.45)
Generally
![r_o=r_1+t](https://tex.z-dn.net/?f=r_o%3Dr_1%2Bt)
![r_o=5.45+0.5](https://tex.z-dn.net/?f=r_o%3D5.45%2B0.5)
![r_o=5.95](https://tex.z-dn.net/?f=r_o%3D5.95)
Generally the equation for outer diameter is mathematically given by
![D_o=2r_o](https://tex.z-dn.net/?f=D_o%3D2r_o)
![D_o=11.9inch](https://tex.z-dn.net/?f=D_o%3D11.9inch)
Therefore
Assuming that the thin cylinder is subjected to integral Pressure
Outer Diameter is
![D_o=11.9inch](https://tex.z-dn.net/?f=D_o%3D11.9inch)