Answer:
The density of the copper is higher than aluminium. Hence it is heavier compared to aluminium conductors it requires strong structures and hardware to bear the weight. More ductile and has high tensile strength.
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Aluminium & Copper properties.
Property Copper (Cu) Aluminium (Al)
Density (g/cm3) 8.96 2.70
Answer:
1+1×1 multiplay then you get the answer
Answer:
Tractive force or traction
Explanation:
The main purpose of the tractive forces is to improve the ability to <u>transform the engine's energy into the vehicle's movement.</u> There are several systems that have different qualities and uses. Here we explain how they work and what they are for.
In a traction vehicle with one of its axles, its ability to transmit engine power to the ground is limited for two reasons:
- At least one of the wheels must have adhesion, and this as long as it has a self-locking differential. Otherwise, simply with one lacking grip, we can no longer move forward.
- If there are two wheels that must distribute the power, it will always be easier to saturate the traction capacity of the tire than if we divide the force by four. The example is very simple: if we try to drag an object on the ground by pulling a rubber, it will stretch more than if we pull 4 identical tires, although the force we make is the same.
Contrary to what one might think, all-wheel drive vehicles are nothing recent. What happens is that it did take time to reduce the weight and size, as well as to increase the resistance of the homokinetic joints (they are articulations on the axles to allow the wheel to go up and down with the suspension or turn with the steering) to to adapt these systems to cars.
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.
I believe it’s A bit I. could be wrong
