Answer:Topographic map. Contour line. Learning Objectives. After completing this chapter, you will be able to: □ Define civil engineering and civil drafting.
Explanation:
Answer:
The answer to this question is 1273885.3 ∅
Explanation:
<em>The first step is to determine the required hydraulic flow rate liquid if working pressure and if a cylinder with a piston diameter of 100 mm is available.</em>
<em>Given that,</em>
<em>The distance = 50mm</em>
<em>The time t =10 seconds</em>
<em>The force F = 10kN</em>
<em>The piston diameter is = 100mm</em>
<em>The pressure = F/A</em>
<em> 10 * 10^3/Δ/Δ </em>
<em> P = 1273885.3503 pa</em>
<em>Then</em>
<em>Power = work/time = Force * distance /time</em>
<em> = 10 * 1000 * 0.050/10</em>
<em>which is =50 watt</em>
<em>Power =∅ΔP</em>
<em>50 = 1273885.3 ∅</em>
Answer:
<u>The automobile rental prices shall show all taxes (including a 6% state tax).</u>
Explanation:
Im pretty sure
Answer:
True
Explanation:
For point in xz plane the stress tensor is given by![\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DDx_%7B%7D%20%26txz%5C%5Ctzx%26Dz%5C%5C%5Cend%7Barray%7D%5Cright%5D)
where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components
We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )
thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane
Answer: 24 pA
Explanation:
As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.
Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.
The resistance R of a given resistor, is expressed by the following formula:
R = ρ L / A
Replacing by the values for resistivity, L and A, we have
R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2
R = 2.1. 10¹¹ Ω
Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:
I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA
