Answer:
B.
Explanation:
For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.
The truth table for a NAND gate with 2 inputs is as follows:
0 0 1
0 1 1
1 0 1
1 1 0
The truth table for an OR gate, is as follows:
0 0 0
0 1 1
1 0 1
1 1 1
If we add two extra columns for inverted inputs, the truth table will be this one:
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0
which is the same as for the NAND gate, not the opposite, so the statement is false.
This means that the right choice is B.
Answer:
I=9.6×e^{-8} A
Explanation:
The magnetic field inside the solenoid.
B=I*500*muy0/0.3=2.1×e ^-3×I.
so the total flux go through the square loop.
B×π×r^2=I×2.1×e^-3π×0.025^2
=4.11×e^-6×I
we have that
(flux)'= -U
so differentiating flux we get
so the inducted emf in the loop.
U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)
so, I=2.9×e^{-6}÷30
I=9.6×e^{-8} A
Answer: 1.137*10^7 Btu/h.
Explanation:
Given data:
Efficiency of the plant = 4.5percent
Net power output of the plant = 150kw
Solution:
The required collection rate
QH = W/n
= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.
= 3333.333 *3412.152Btu/h.
= 11373840 Btu/h
= 1.137*10^7 Btu/h.
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.
