Answer:
minimum length of a surface crack is 18.3 mm
Explanation:
Given data
plane strain fracture toughness K = 82.4 MPa m1/2
stress σ = 345 MPa
Y = 1
to find out
the minimum length of a surface crack
solution
we will calculate length by this formula
length = 1/π ( K / σ Y)²
put all value
length = 1/π ( K / σ Y)²
length = 1/π ( 82.4 
/ 345× 1)²
length = 18.3 mm
minimum length of a surface crack is 18.3 mm
Answer:
168.57 mV
Explanation:
Initial magnetic flux = BA , B magnetic field and A is area of loop
= .35 x 3.14 x .37²
= .15 Weber
Final magnetic flux
= - .2 x 3.14 x .37²
= - .086 Weber
change in flux
.15 + .086
= .236 Weber
rate of change of flux
= .236 / 1.4
= .16857 V
= 168.57 mV
Answer:
Explanation:
We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector
D₁ = - 255 cos 49 i + 255 sin49 j
= - 167.29 i + 192.45 j
Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is
D = 125 i
So
D₁ + D₂ = D
- 167.29 i + 192.45 j + D₂ = 125 i
D₂ = 125 i + 167.29 i - 192.45 j
= 292.29 i - 192.45 j
Angle of D₂ with x axes θ
tan θ = -192.45 / 292.29
= - 0.658
θ = 33.33 south of east
Magnitude of D₂
D₂² = ( 192.45)² + ( 292.29)²
D₂ = 350 km approx
Tan
Answer:
a = F / m
Explanation:
force same -> mass variable
more mass -> less force
