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Greeley [361]
3 years ago
15

What is the damped natural frequency (in rad/s) of a second order system whose undamped natural frequency is 25 rad/s and has a

damping ratio of 0.3?
Engineering
1 answer:
TiliK225 [7]3 years ago
3 0

Answer:

damped natural frequency = 23.84 rad/s

Explanation:

given data

damping ratio = 0.3

undamped natural frequency = 25 rad/s

to find out

damped natural frequency of a second order system

solution

we know that if damping ratio is = 0

then it is undamped system

and if damping ratio is > 1

then it is overdamped system

and  and if damping ratio is ≈ 1

then it is critical damped system

so damped natural frequency of a second order system formula is

damped natural frequency = wn × \sqrt{1-r^2}

here wn is undamped natural frequency and r is damping ratio

damped natural frequency = 25 × \sqrt{1-0.3^2}

damped natural frequency = 23.84 rad/s

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marishachu [46]

Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

Rz = 50 N

Sum of moments about the x axis:

∑τx = Iα

Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0

Mx = 22 Nm

Sum of moments about the y axis:

∑τy = Iα

My = 0 Nm

Sum of moments about the z axis:

∑τz = Iα

Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0

Mz = -30.8 Nm

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4 years ago
A gas stream contains 4.0 mol % NH3 and its ammonia content is reduced to 0.5 mol % in a packed absorption tower at 293 K and 10
bagirrra123 [75]

Answer:

Explanation:

Step by step solved solution is given in the attached document.

8 0
3 years ago
Two balls are chosen randomly from an urn containing 8 white 4 black, and orange balls. Suppose that we win $ 2 for each black b
Scorpion4ik [409]
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A heat engine that rejects waste heat to a sink at 520 R has a thermal efficiency of 35 percent and a second- law efficiency of
xeze [42]

Answer:

The source temperature is 1248 R.

Explanation:

Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.

Given:  

Temperature of the heat sink is 520 R.

Second law efficiency is 60%.

Actual thermal efficiency is 35%.

Calculation:  

Step1

Reversible efficiency is calculated as follows:

\eta_{II}=\frac{\eta_{a}}{\eta_{rev}}

0.6=\frac{0.35}{\eta_{rev}}

\eta_{rev}=0.5834

Step2

Source temperature is calculated as follows:

\eta_{rev}=1-\frac{T_{L}}{T}

\eta_{rev}=1-\frac{520}{T}

0.5834=1-\frac{520}{T}

T = 1248 R.

The heat engine is shown below:

Thus, the source temperature is 1248 R.

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Explanation:

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