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3 years ago

What is an example of a mechanical noise?

1 answer:

4 0

Resonance:
The resounding recurrence is the recurrence at which a bit of metal, plastic or whatever else swings/vibrates with minimal measure of vitality input. Think about a man on a play area swing. You realize that it requires next to no push to keep the individual swinging. The recurrence at which they swing forward and backward is their full recurrence. In the event that you endeavor to influence them to swing speedier or slower, it will take altogether more vitality.

Resonating Panels:
This kind of clamor is caused when the bass notes are an indistinguishable recurrence from the thunderous recurrence of a metal or plastic board. To stop or decrease the commotion related with this kind of issue, you can do two or three things.

Rattling:
This sort of commotion would be caused when 2 bits of metal, plastic, whatever... are sufficiently close to hammer into each other when they resound. This is most likely best illuminated by filling the hole between the two vibrating parts with silicone sealant or shut cell froth climate stripping. The climate stripping is a superior arrangement in places like behind the tag. On the off chance that you have a tag outline, you can get some truly thin climate stripping and put between the casing and the plate.

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What is the IEEE 802.11a center frequency in GHz, and what are its max/min data rates in Mbps? What is the IEEE 802.11b center f

kogti [31]

Explanation:

What is IEEE 802.11?

IEEE 802.11 is a set of WLAN standards for communication developed by the Institute for Electrical and Electronics Engineers (IEEE) and is unarguably most widely used WLAN technology.

Features: IEEE 802.11a

The operating frequency band is 5 GHz.
The maximum theoretical data rate is 54 Mbps, the typical throughput is around 25 Mbps and minimum data rate is 6 Mbps.
It can support 64 users per access point.

Features: IEEE 802.11b

The operating frequency band is 2.4 GHz.
The maximum theoretical data rate is 11 Mbps but typical throughput is around 6 Mbps and minimum data rate is 1 Mbps.
It can support 32 users per access point.

Wireless Coverage IEEE 802.11a Vs IEEE 802.11b:

Signal coverage is one of the most important factors among users.
The transmission range of IEEE 802.11a is not greater than 100 ft in indoor setting whereas IEEE 802.11b has a superior performance in this regard with transmission range up to 150 ft in indoor setting.
The data rate has a direct relation with the access point coverage area, a higher data rate means less coverage area and a lower data rate results in increased coverage.

8 0

3 years ago

So... if bedbugs live in beds, where do cockroaches live?

Deffense [45]

Answer:

oo.p i wish I could answer that

Explanation:

6 0

3 years ago

Read 2 more answers

I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f

cricket20 [7]

Answer:

$a=4.44\frac{m}{s^2}$

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

$t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s$

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

$d=vt\\d=20\frac{m}{s}(0.5s)=10m$

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

$d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}$

7 0

3 years ago

Total these measurements. Your answer should indicate the proper accuracy. Be sure to include the units in your answer. (Remembe

vivado [14]

Answer:

Explanation:

This is tough. The last number 0.2 has only one significant figure. So while the sum of all the numbers is 12.3, you must only leave one sig figure. Rounding to the tenths gives 10.

3 0

3 years ago

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert

cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, $k=4.2\times 10^5\ N/m$

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

$mgh=\dfrac{1}{2}kx^2$

x is the compression in spring

$x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m$

So, the compression in the spring is 5.88 meters.

6 0

3 years ago