The answer 1 is
The best-known use of radio waves is for communication; television, cellphones and radios all receive radio waves and convert them to mechanical vibrations in the speaker to create sound waves that can be heard.Electromagnetic radiation is transmitted inwaves or particles at different wavelengths and frequencies.
the answer 2 is
Radio Waves: Instant Communication
Microwaves: Data and Heat
i think the answer number 3 is
It is all about wavelength versus tunnel diameter. The wavelength of GPS is about 20cm it would happily propagate in any normal tunnel if it could get in but the earth and other structures absorb it. AM radio (600kHz - 1500kHz) cannot propagate in any normal tunnel because the wavelength is too long (500m-200m) relative to the diameter, and thus gets reflected at the entrance. FM (100MHz ~ 3m) would propagate and it does for a while but then it suffers reflections
Ans: Time <span>taken by a pulse to travel from one support to the other
= 0.348s</span>
Explanation:First you need to find out the speed of the wave.
Since
Speed = v =

Where
T = Tension in the cord = 150N
μ = Mass per unit length = mass/Length = 0.65/28 = 0.0232 kg/m
So
v =

= 80.41 m/s
Now the time-taken by the wave = t = Length/speed = 28/80.41=
0.348s
Answer:

Explanation:
We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.
1 m =100 cm
Surface area =S=


We have to find the charge Q on the positive plates of the capacitor.
V=Initial voltage between plates
d=Initial distance between plates
Initial Capacitance of capacitor

Capacitance of capacitor after moving plates


Potential difference between plates after moving








Hence, the charge on positive plate of capacitor=
Take a look at a simple reaction like the one below:
In this reaction some reactant A is turned into some product B. The rate of reaction can be represented by a decrease in concentration of A over time or as the increase of B over time. This is written: