Answer:
<em>a) 4.51 lbf-s^2/ft</em>
<em>b) 65.8 kg</em>
<em>c) 645 N</em>
<em>d) 23.8 lb</em>
<em>e) 65.8 kg</em>
<em></em>
Explanation:
Weight of the man on Earth = 145 lb
a) Mass in slug is...
32.174 pound = 1 slug
145 pound = 
slug
= 145/32.174 = <em>4.51 lbf-s^2/ft</em>
b) Mass in kg is...
2.205 pounds = 1 kg
145 pounds = kg
= 145/2.205 = <em>65.8 kg</em>
c) Weight in Newton = mg
where
m is mass in kg
g is acceleration due to gravity on Earth = 9.81 m/s^2
Weight in Newton = 65.8 x 9.81 = <em>645 N</em>
d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,
1 m/s^2 = 3.2808 ft/s^2
m/s^2 = 5.30 ft/s^2
= 5.30/3.2808 = 1.6155 m/s^2
weight in Newton = mg = 65.8 x 1.6155 = 106
weight in pounds = 106/4.448 = <em>23.8 lb</em>
e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth
mass on the moon = <em>65.8 kg</em>
Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.
Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6 
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
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))
