Answer:
“I really love the design work in engineering, the face-to-face interaction with clients, and the opportunity to see projects come to life. But if I had to pick one thing that I don’t enjoy as much, I would have to say it’s contract preparation.”
Answer:
a) 0.3
b) 3.6 mm
Explanation:
Given
Length of the pads, l = 200 mm = 0.2 m
Width of the pads, b = 150 mm = 0.15 m
Thickness of the pads, t = 12 mm = 0.012 m
Force on the rubber, P = 15 kN
Shear modulus on the rubber, G = 830 GPa
The average shear strain can be gotten by
τ(average) = (P/2) / bl
τ(average) = (15/2) / (0.15 * 0.2)
τ(average) = 7.5 / 0.03
τ(average) = 250 kPa
γ(average) = τ(average) / G
γ(average) = 250 kPa / 830 kPa
γ(average) = 0.3
horizontal displacement,
δ = γ(average) * t
δ = 0.3 * 12
δ = 3.6 mm
Answer:
Their company address
When their last tax period started
How often they have to file a tax return
When they started collecting sales tax for the agency
Explanation:
For setup the sales tax information in Quickbooks online for a client who only does business in their home state, we need these information which are given below:
1. Their company addresses
2. Last Tax Time period started
3. How frequently they filed the tax return
3. when they begin to received sales tax
Therefore all the other options are not valid. Hence, ignored it
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:
A fluid is a substance [ <em>liquid</em><em> </em><em>and</em><em> </em><em>gas</em><em> </em><em>state</em><em> </em>] in which motion of another substance in it is opposed due to viscous drag [ <em>viscosity</em><em> </em>]
Explanation:
