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
Λ = 3*10^8 / 9*10^8 = 1/3 m
no. of wavelengths = 60/(1/3) = 180
Answer:
F = 19.1 N
Explanation:
To find the force exerted by the string on the block you use the following formula:
(1)
k: spring constant = 95.5 N/m
x: displacement of the block from its equilibrium position = 0.200 m
you replace the values of k and x in the equation (1):
Hence, the force exterted on the block is 19.1 N
Answer:
D. A lower Voltage into a higher
An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in an ionised gas (plasma).
