Answer:
a)Yes will deform plastically
b) Will NOT experience necking
Explanation:
Given:
- Applied Force F = 850 lb
- Diameter of wire D = 0.15 in
- Yield Strength Y=45,000 psi
- Ultimate Tensile strength U = 55,000 psi
Find:
a) Whether there will be plastic deformation
b) Whether there will be necking.
Solution:
Assuming a constant Force F, the stress in the wire will be:
stress = F / Area
Area = pi*D^2 / 4
Area = pi*0.15^2 / 4 = 0.0176715 in^2
stress = 850 / 0.0176715
stress = 48,100.16 psi
Yield Strength < Applied stress > Ultimate Tensile strength
45,000 < 48,100 < 55,000
Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.
Answer:

Explanation:
Note that acceleration is the rate change of velocity i.e
.
Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have


Answer:
Starts on Saturday, June 1
and ends on
Saturday, November 30
Explanation:
Answer:
The correct option is: Total energy
Explanation:
The Hamiltonian operator, in quantum mechanics, is an operator that is associated with the<u> total energy of the system.</u> It is equal to the sum of the total kinetic energy and the potential energy of all the particles of the system.
The Hamiltonian operator was named after the Irish mathematician, William Rowan Hamiltonis denoted and is denoted by H.
Answer is c, they are equal:
Explanation: