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:
15 meters
Explanation:
The inicial energy of the ball is just potencial energy, and its value is:
E = m * g * h = m * g * 20,
where m is the ball mass, and g is the value of gravity.
In the moment that the ball strickes the ground, all potencial energy transformed into kinetic energy, and 25% of this energy is lost, so the total energy at this moment will be:
E' = 0.75 * E = 0.75 * m * g * 20 = 15*m*g
This kinetic energy will make the ball goes up again, and at the maximum height, all kinetic energy is transformed back into potencial energy.
So, as the mass and the gravity are constants, we can calculate the height the ball will reach:
E' = m*g*h = 15*m*g -> h = 15 meters
I haven´t learned that yet I´m sorry.
Closer u get 2 the center the more balanced out your weight will b
