Answer:
This doesn't represent an equilibrium state of stress
Explanation:
∝ = 1 , β = 1 , y = 1
x = 0 , y = 0 , z = 0 ( body forces given as 0 )
Attached is the detailed solution is and also the conditions for equilibrium
for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution
Answer:
8 for dual-op-amp package, and 14 for quad-op-amp
Explanation;
This is because every op-amp has 2 input terminal 4 pns
So one output terminal that is 2 pins which are required for power
and the same for a minumum number of pins required by quad op amp which is 14
The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.
<h3>How to determine the amount of settlement?</h3>
For a layer of 3.8 m thickness, we were given the following parameters:
U = 50% = 0.5.
Sc = 7.3 cm.
For Sf, we have:
Sf = Sc/U
Sf = 7.3/0.5
Sf = 14.6
Therefore, Sf for a layer of 38 m thickness is given by:
Sf = 14.6 × 38/3.8
Sf = 146 cm.
At 50%, the time for a layer of 3.8 m thickness is:
= 1.5 year.
At 50%, the time for a layer of 38 m thickness is:
= 1.5 × (38/3.8)²
= 150 years.
For the thickness of 38 m, U₂ is given by:
![\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25 \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05](https://tex.z-dn.net/?f=%5Cfrac%7BU_1%5E2%7D%7BU_2%5E2%7D%20%3D%5Cfrac%7B%28T_v%29_1%7D%7B%28T_v%29_2%7D%20%3D%20%5Cfrac%7Bt_1%7D%7Bt_2%7D%20%5C%5C%5C%5CU_2%5E2%20%3D%20U_1%5E2%20%5Ctimes%20%5B%5Cfrac%7Bt_2%7D%7Bt_1%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.5%5E2%20%5Ctimes%20%5B%5Cfrac%7B1.5%7D%7B150%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.25%20%20%5Ctimes%200.01%5C%5C%5C%5CU_2%3D%5Csqrt%7B0.0025%7D%20%5C%5C%5C%5CU_2%3D0.05)
The new settlement after 1.5 year is:
Sc = U₂Sf
Sc = 0.05 × 146
Sc = 7.3 cm.
For time, t₂ = 5 year:
![U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25 \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09](https://tex.z-dn.net/?f=U_2%5E2%20%3D%20U_1%5E2%20%5Ctimes%20%5B%5Cfrac%7Bt_2%7D%7Bt_1%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.5%5E2%20%5Ctimes%20%5B%5Cfrac%7B5%7D%7B150%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.25%20%20%5Ctimes%200.03%5C%5C%5C%5CU_2%3D%5Csqrt%7B0.0075%7D%20%5C%5C%5C%5CU_2%3D0.09)
The new settlement after 5 year is:
Sc = U₂Sf
Sc = 0.09 × 146
Sc = 13.14 cm.
Read more on clay layer here: brainly.com/question/22238205
Answer:
critical stress
= 1382.67 MPa
Explanation:
given data
plane strain fracture toughness = 54.8 MP
length of surface creak = 0.5 mm
we take here
parameter Y = 1.0
solution
we apply critical stress formula that is
critical stress
=
.............................1
here K is design stress plane strain fracture toughness and a is length of surface creak so put all these value in equation 1
critical stress
=
solve it we get
critical stress = 1382.67 MPa
As exposed stress 1030 MPa is less than critical stress 1382 MPa
so that fracture will not be occur here
Answer:
a. 10bits
b. 7 bits
c. 6 bits
Explanation:
a. for 0 to 512
# of numbers = 512 - 0 + 1 =513
[log ₂513] = 9 bits
we actually need 10 bits
b. for 0 to 75
# of numbers = 75 - 0 + 1 =76
[log ₂76] = 6 bits
we actually need 7 bits
c. for -20 to 13
# of numbers = 13 - (-20) + 1 =34
[log ₂34] = 5 bits
we actually need 6 bits