I put
People who pursue a career in the creative imaging fields have qualities like a good imagination, creativity, open minds, good with ideas, and handling situations. If you enter that field, you need imagination to create things and an open mind to be open to all creations. You need good ideas to make good thing that will work.
please don't plagiarise tho, re-word it.
Answer:
I think it's 23 ohms.
Explanation:
Not entirely sure about it.
hope this helps
Answer:
the velocity = 10 m / sec if an object moves 100 m in 10s
Answer: The overhead percentage is 7.7%.
Explanation:
We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.
We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.
So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:
OH % = (58 / 758) * 100 = 7.7 %
Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
