Answer:
for 1st question the answer is 5th option.
for 2nd question the answer is 2nd option
hope it helps you mate
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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
Answer:
The answer is copying a program to give to someone else to use
and burning a copy of a DVD to sell
Explanation:
Software piracy is the act of stealing software that is legally protected. This stealing includes copying, distributing, modifying or selling
Answer:
0.024 m = 24.07 mm
Explanation:
1) Notation
= tensile stress = 200 Mpa
= plane strain fracture toughness= 55 Mpa
= length of a surface crack (Variable of interest)
2) Definition and Formulas
The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.
By definition we have the following formula for the tensile stress:
(1)
We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for
Multiplying both sides of equation (1) by 
(2)
Sequaring both sides of equation (2):
(3)
Dividing both sides by 
we got:
![\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5B%5Cfrac%7BK%7D%7BY%5Csigma_c%7D%5D%5E2)
(4)
Replacing the values into equation (4) we got:
![\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5B%5Cfrac%7B55%20Mpa%5Csqrt%7Bm%7D%7D%7B1.0%28200Mpa%29%7D%5D%5E2%20%3D0.02407m)
3) Final solution
So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.
