Answer:
310 meters
Explanation:
Given:
v₀ = 0 m/s
t = 8.0 s
a = -9.8 m/s²
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (8.0 s) + ½ (-9.8 m/s²) (8.0 s)²
Δy = -313.6
Rounded to two significant figures, the object fell 310 meters.
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:
Explanation:
Q1 = 35 nC = 35 x 10^-9 C
m = 3.5 micro gram = 3.5 x 10^-9 Kg
d = 35 cm = 0.35 m
(a) The electrostatic force between the two charges is balanced by the weight of another charge.
F = m g
(b) By substituting the values
Q2 = 13.34 x 10^-12 C
Q2 = 0.0134 nC
The gravitational pull is weaker.
