Answer:

Explanation:
First I sketched the compression of the gas with the help of the given pressure change process relation. That is your pressure change due to change in volume.
To find the area underneath the curve (the same as saying to find the work done) you should integrate the given relation for pressure change:

Answer:
Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact disc reader. Since you did such a good job in your design, the company decided to work with you in their latest Blue-ray readers, as well. However, this time the task is that once the user hits eject button, the motor that spins the disc slows down from 2000 rpm to 300 rpm and at 300 rpm a passive torsional spring-damper mechanism engages to decelerate and stop the disc. Here, your task is to design this spring-damper system such that the disc comes to rest without any oscillations. The rotational inertia of the disc (J) is 2.5 x 10-5kg m² and the torsional spring constant (k) is 5 × 10¬³NM. Calculate the critical damping coefficient cc for the system. choice of the damper, bear in mind that a good engineer stays at least a factor of In your 2 away from the danger zone (i.e., oscillations in this case). Use the Runge Kutta method to simulate the time dependent angular position of the disc, using the value of damping coefficient (c) that calculated. you Figure 1: Blue-ray disc and torsional spring-damper system.
Answer: Create lessons learned at the end of the project.
Explanation:
Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.
The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.
Answer:
W = 11,416.6879 N
L ≈ 64.417 cm
Explanation:
The maximum shear stress,
, is given by the following formula;

= 1 cm = 0.01
h = 29 cm = 0.29 m
= 25 cm = 0.25 m
b = 15 cm = 0.15 m
= The centroidal moment of inertia
= 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴
Substituting the known values gives;

= 1.2257083
× 10⁻⁴ m⁴
From which we have;

Which gives;
W = 11,416.6879 N

= 1500 N/cm² = 15,000,000 N/m²
= 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²
From Which we have;


L ≈ 0.64417 m ≈ 64.417 cm.
Answer:
The interior of the metal wire is neutral
The net electric field everywhere inside the wire is zero
There may be excess charges on the surface of the wire
There is no net flow of mobile electrons inside the wire
There are no excess charges in the interior of the wire
Explanation:
When a metal wire is in equilibrium, the interior of the metal wire is always neural and there is no net flow of mobile electrons inside the wire. Additionally, there are no excess charges in the interior of the wire but there is likelihood of having excess charges on the surface of the wire. Finally, the net electric field everywhere inside the wire is zero