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kumpel [21]
3 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]3 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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An electron and a second particle both move in circles perpendicular to a uniformmagnetic field. The mass of the second particle
Katarina [22]

Answer:

The change on the second particle is 2.93\times 10^{-16}\ C.

Explanation:

The period of revolution of the particle in the magnetic field is given by the formula as follows :

T=\dfrac{2\pi m}{Bq}

It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.

m_s=m_p

If both particles take the same amount of time to go once around their respective circles. So,

T_e=T_s\\\\\dfrac{2\pi m_e}{Bq_e}=\dfrac{2\pi m_s}{Bq_s}\\\\\dfrac{m_e}{q_e}=\dfrac{m_p}{q_s}\\\\q_s=\dfrac{m_pq_e}{m_e}\\\\q_s=\dfrac{1.67\times 10^{-27}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}\\\\q_s=2.93\times 10^{-16}\ C

So, the change on the second particle is 2.93\times 10^{-16}\ C.

7 0
3 years ago
A 2kg Book is sitting on a table.a 10n Force is pulling to the right.a 3n Force is pulling to the left what is the net force act
salantis [7]


The net force = sum of all forces acting on the body



If we take left side as -ve and right side as +ve,
then,

The net force here would be equal to,
10N + (- 3N)
= 7N.


Therefore, a net force of +7N ( + indicates it's moving towards right) is acting on the book of mass 2kg.


4 0
3 years ago
Question #4
IrinaVladis [17]
2) transverse



hope this helped:)
3 0
3 years ago
Read 2 more answers
một quả bóng quần vợt nặng 0,060kg chuyển động với tốc độ 45,0m/s va vào bức tường dưới một góc 45 độ rồi bật trở ra với ucngf t
Vinvika [58]

can you tell in English.......

5 0
3 years ago
A woman and her dog are out for a morning run to the river, which is located 4.0 KM away. The woman runs at 2.5 M/S in a straigh
VLD [36.1K]

River shore is located at distance

d = 4 km

speed of the woman is given as

v_1 = 2.5 m/s

now the time taken by the woman to cover the distance is

t = \frac{d}{v}

t = \frac{4000}{2.5} = 1600 s

for the same time interval the dog will run to and fro with speed 4.5 m/s

so the total distance moved by the dog is given by

d = v* t

d = 4.5 * 1600

d = 7200 m

<em>so the total distance that dog will move is 7200 m or 7.2 km</em>

8 0
4 years ago
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