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:
11.48 m
Explanation:
A brick starts from rest and gains a speed of 15 m/s and accelerates at 9.8 m/s^2
u = 0
v= 15
a= 9.8
s= ?
V^2 = U^2 + 2as
15^2 = 0^2 + 2 × 9.8×s
225= 19.6s
s= 225/19.6
s = 11.48m
Hence the brick will fall 11.48 m
Answer:
A driver.
Explanation:
Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.
Answer:
The distance traveled during its acceleration, d = 214.38 m
Explanation:
Given,
The object's acceleration, a = -6.8 m/s²
The initial speed of the object, u = 54 m/s
The final speed of the object, v = 0
The acceleration of the object is given by the formula,
a = (v - u) / t m/s²
∴ t = (v - u) / a
= (0 - 54) / (-6.8)
= 7.94 s
The average velocity of the object,
V = (54 + 0)/2
= 27 m/s
The displacement of the object,
d = V x t meter
= 27 x 7.94
= 214.38 m
Hence, the distance the object traveled during that acceleration is, a = 214.38 m
Answer:
Because they are both easy to measure (?)
Explanation:
(I'm not really sure, there are no choices. If there were different options I might be able to better answer this)
