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kumpel [21]
2 years ago
6

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

requency in GHz, and what are its max/min data rates in Mbps? Assuming other factors are the same, which one has larger wireless coverage (a or b), and why? (1 point)
Physics
1 answer:
kogti [31]2 years ago
8 0

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.
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Canola oil is less dense than water, so it floats on water, but its index of refraction is 1.47, higher than that of water. When
kupik [55]

Answer:

therefore critical angle c= 69.79°

Explanation:

Canola oil is less dense than water, so it floats over water.

Given n_{canola}= 1.47

which is higher than that of water

refractive index of water n_{water}=1.33

to calculate critical angle of light going from the oil into water

we know that

sinc= \frac{n_{water}}{n_{canola}}

now putting values we get

sinc= \frac{1.33}{1.47}

c= sin^{-1}(\frac{1.33}{1.47} )

c=69.79°

therefore critical angle c= 69.79°

8 0
3 years ago
A spring-mass system has a spring constant k = 383 N/m, length of the spring L = 0.5 m, and the mass attached to it is M = 3.8 k
sergiy2304 [10]

Answer:

Frequency = f = 10.0394 (1/s)

Explanation:

The frequency of oscillation of the system is given by the action:

f= √(k/m)

f= system count

k = spring constant

m = mass connected to the spring

Therefore the frequency will be:

f= √(k/m) = √(383(N/m) / (3.8kg))= √( 100.7895 (kg×m/s²)/(kg ) =

= √( 100.7895 (1/s²) = 10.0394 (1/s)

4 0
2 years ago
When a piece of metal mass of 72.17 g is
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Answer:

Explanation:

The trick is in finding the volume.

Final Volume = 26.64

Initial Volume=<u>20.92</u>                   Subtract

Metal Volume  5.72  cm^3

Density = mass / volume

Density = 72.17 / 5.72

Density = 12.617

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2 years ago
Can we use a clinical thermometer to measure the temperature of a candle flame​
miv72 [106K]

In theory, yes. The 2 problems are the materials used for clinical thermometers, & the temperature capacity of the clinical thermometer. If anything, change the material & extend the measurement threshold. At that point, it wouldn´t be used for clinical garbage anymore.

3 0
3 years ago
Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo
ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

a)

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

F_{net}=ma

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

a=9.50 m/s^2 (acceleration)

So, the net force is

F_{net}=(94.0)(9.50)=893 N

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

b)

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

v=u+at

where

v is the  final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

a=9.50 m/s^2 (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

v=0+(9.50)(0.890)=8.5 m/s

c)

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

W=K_f - K_i = \frac{1}{2}mv^2-0

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

K_i = 0 J is the initial kinetic energy

So the work done is

W=\frac{1}{2}(94.0)(8.5)^2=3396 J

The power expended is given by

P=\frac{W}{t}

where

t = 0.890 s is the time elapsed

Substituting,

P=\frac{3396}{0.890}=3816 W

d)

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

F_1=893 N

We know that the average force for the whole race is

F_{avg}=820 N

Which can be rewritten as

F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}

And solving for F_2, we find the average force exerted by Bolt on the ground during the second phase:

F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N

The net force exerted by Bolt during the second phase can be written as

F_{net}=F_2-D (1)

where D is the air drag.

The net force can also be rewritten as

F_{net}=ma

where

a=\frac{v-u}{t} is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

a=\frac{12.4-8.5}{8.69}=0.45 m/s^2

So we can now find the average drag force from (1):

D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

\Delta E=W=Ds

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m

And so,

\Delta E=Ds=(770.2)(90.6)=69780 J

e)

The power that Bolt must expend just to voercome the drag force is given by

P=\frac{\Delta E}{t}

where

\Delta E is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

\Delta E=69780 J

t = 8.69 s is the time elapsed

Substituting,

P=\frac{69780}{8.69}=8030 W

And we see that it is about twice larger than the power calculated in part c.

3 0
3 years ago
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