Answer:
The energy released during nucleur fissionor fusion , espicially when used to generate
Explanation:
Distionary.
<h3>Answer : </h3><h3 /><h3>A ) The larger gear can be moved by applying a relatively small force on the smaller gear.</h3>
<h3>B )
The force applied on the smaller gear is transmitted without any loss to the larger gear .</h3><h3 /><h3>
C ) the direction of motion can be changed without changing the direction of the applied force .</h3>
D ) the system would continue to move without any further, after and initial force has set in motion.
Linear momentum is the product of mass and velocity. In this case, it is simply:
Answer:
Explanation:
We need the power equation here:
P = W/t where W is work and is defined as
W = F*displacement.
Force is a measure in Newtons, which is also weight. We have the mass of the piano, but we need to find the weight:
w = mg so
w = 166(9.8) so
w = 1600N, rounded to the correct number of sig dig. We use that now in the power equation:
and isolating the unknown:
so
t = 5.3 seconds
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
