Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa
<u>Explanation:</u>
Given -
Stress Direction, A = [1 0 0 ]
Slip plane = [ 1 1 1]
Normal to slip plane, B = [ 1 1 1 ]
Critical stress, Sc = 2.92 MPa
Let the direction of slip on = [ 1 1 0 ]
Let Ф be the angle between A and B
cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3
cos Ф = 1/√3
σ = Sc / cosФ cosλ
For slip along [ 1 1 0 ]
cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1
cos λ = 1/√2
Therefore,
σ = 2.92 / 1/√3 1/√2
σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa
Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa
Answer:
Of longitudinal waves
Explanation:
Depending on the direction of the oscillation, there are two types of waves:
- Transverse waves: in a transverse wave, the oscillations occur perpendicularly to the direction of propagation of the wave. Examples are electromagnetic waves.
- Longitudinal waves: in a longitudinal wave, the oscillations occur parallel to the direction of propagation of the wave. In such a wave, the oscillations are produced by alternating regions of higher density of particles, called compressions, and regions of lower density of particles, called rarefactions. Examples of longitudinal waves are sound waves.
As the metal expands as does the road bed so neither really effevts those foing over the bridge. as it is hot the metal will expand and so will most tarmac on roads.
Answer:
0.65 kg*m/s and 0.165 kg*m/s
Explanation:
Step one:
given data
mass m= 0.5kg
initial velolcity u=1.3m/s
final velocity v= 0.97m/s
Required
The change in momentum
Step two:
We know that the expression for impulse is given as
Ft= mv
Ft= 0.5*1.3
Ft= 0.65 kg*m/s
The expression for the change in momentum is given as
P= mΔv
substitute
Pt= 0.5*(1.3-0.97)
Pt= 0.5*0.33
Pt=0.165 kg*m/s
