C) above, the Earth's core is about the melting point of iron.
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.
To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

Here,
M = Mass
R = Radius of the hoop
The precession frequency is given as

Here,
M = Mass
g= Acceleration due to gravity
d = Distance of center of mass from pivot
I = Moment of inertia
= Angular velocity
Replacing the value for moment of inertia


The value for our angular velocity is not in SI, then


Replacing our values we have that


The precession frequency is




Therefore the precession period is 5.4s
Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C
It shortens so that the tips reach faster