Answer:
w=2.25
Explanation:
It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.
The surface of the cross-section of the stapes was determined:
A_ab= 10 mm^2
A-cd= 15 mm^2
The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.
σ_ab = F_ab/A_ab
σ_allow
σ_cd = F_cd/A_cd
σ_allow
In the next step we will determine the static size: Picture b).
We apply the conditions of equilibrium:
∑F_x=0
∑F_y=0
∑M=0
∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0
==> F_cd = 2*w*k*N
∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0
==> F_ab = w*k*N
Now we determine the load w
<u>Sector AB: </u>
σ_ab = F_ab/A_ab
σ_allow=300 KPa
= w/10*10^-6
σ_allow=300 KPa
w_ab = 3*10^-3 kN/m
<u>Sector CD: </u>
σ_cd = F_cd/A_cd
σ_allow=300 KPa
= 2*w/15*10^-6
σ_allow=300 KPa
w_cd = 2.25*10^-3 kN/m
w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}
==> w=2.25 * 10^-3 kN/m
<u>The solution is: </u>
w=2.25 N/m
note:
find the attached graph
Answer:
Explanation:
Using equation of pure torsion

where
T is the applied Torque
is polar moment of inertia of the shaft
t is the shear stress at a distance r from the center
r is distance from center
For a shaft with
Outer Diameter
Inner Diameter

Applying values in the above equation we get
x 
Thus from the equation of torsion we get

Applying values we get

T =829.97Nm
Gas turbines extract the energy from combustion gas
and heat engines convert thermal and chemical energy to mechanical energy
gas is considered a chemical so i'm pretty sure it is considered one
Given:
Wall's inside temperature, 
Room air temperature, 
Solution:
To calculate percentage max relative humidity, we make use of steam table for saturated pressure of wall and air:
From steam table:
At
:

At
:

Now,


(RH)_{max} = 68.58%
Therefore, max Relative Humidity of the air before the occurrence of condensation in wall is 68.85%
Answer:
V=1601gal
Explanation:
Hello! This problem is solved as follows,
First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.
This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.
P=Poil+Patm
P=total pressure or absolute pressure=26psi=179213.28Pa
Patm= the atmospheric pressure =101325Pa
Poil=pressure due to the weight of olive oil=0.86αgh
α=density of water=1000kg/m^3
g=gravity=9.81m/s^2
h= height that olive oil reaches
solving
P=Poil+Patm
P=Patm+0.86αgh
[/tex]
Now we can use the equation that defines the volume of a cylinder.
V=
D=3ft=0.9144m
h=9.23m
solving

finally we use conversion factors to find the volume in gallons
