Answer:
shear strength = 2682.31 Ib/ft^2
Explanation:
major principal stress = 100 Ib / in2
minor principal stress = 20 Ib/in2
Normal stress = 3000 Ib/ft2
<u>Determine the shear strength when direct shear test is performed </u>
To resolve this we will apply the coulomb failure criteria relationship between major and minor principal stress a
for direct shear test
use Mohr Coulomb criteria relation between normal stress and shear stress
Shear strength when normal strength is 3000 Ib/ft = 2682.31 Ib/ft^2
attached below is the detailed solution
Answer:
Both come from the sun
Both are reusable sources
and both don't cause pollution
Explanation:
Explanation:
1) Work done = force x distance x cos(θ)
= 0.15 x 6 x cos(30)
= 0.779
2) Ek = ½mv²
v = acceleration due to gravity so 9.81
Ek = ½(2)(9.81)²
Ek = 96.2361
3) v = (√(2em)) / m
= (√(2(96.2361)(2)) / 2
= 9.81 so especially with no time given, I can only assume the acceleration due to gravity but take it with a pinch of salt.
Answer:
6.004 Ω
Explanation:
For a Y- connected system given that :
Line voltage, 
Line current , 
and specified that 
are in phase.
Hence the impedance will be pure resistive.
For Y-system
= phase voltage
= 120.08 V
Line current = Phase current
Now, 
= 6.004 Ω
