Answer:
2.455 W
Explanation:
The power dissipated in each branch is ...
P = V^2/R
So, the branch powers are ...
branch 1: 18^2/220 ≈ 1.473 W
branch 2: 18^2/330 ≈ 0.982 W
Total power is ...
1.473 W + 0.982 W = 2.455 W
Answer:
The natural angular frequency of the rod is 53.56 rad/sec
Explanation:
Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure
We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

Hence we can write

Comparing with the standard spring equation
we find the cantilever analogous to spring with 
Now the angular frequency of a spring is given by
where
'm' is the mass of the load
Thus applying values we get

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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:
The total load carried by the fiber will be "98%".
Explanation:
The given values are:






, 
As we know,
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
Now,
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
Therefore,
The load carried by fiber,


i.e., 98%