A radio station broadcast on a frequency of 3.7 mhz what is the energy of the radio wave A radio station broadcasts its programmes at a wavelength of 500 m. Find the frequency of the radiowaves transmitted by the radio station, if the speed of radiowaves in air is 3 x 108 m/s. Ans: 6 x 10 Hz
<h3>What is
radio station ?</h3>
Radio broadcasting is the act of sending audio (sound), occasionally together with accompanying metadata, across radio waves to radio receivers used by the general public. Unlike satellite radio, which uses a satellite in Earth's orbit, terrestrial radio broadcasting uses a land-based radio station to transmit radio waves. The listener needs a broadcast radio receiver to hear the material (radio). A radio network with which stations frequently have affiliations provide content in a standard radio format, whether through broadcast syndication, simulcasting, or both. Radio stations use a variety of modulations to transmit their signals, including FM (frequency modulation), which is an older analog audio standard, and AM (amplitude modulation).
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Answer:
D) momentum of cannon + momentum of projectile= 0
Explanation:
The law of conservation of momentum states that the total momentum of an isolated system is constant.
In this case, the system cannon+projectile can be considered as isolated, because no external forces act on it (in fact, the surface is frictionless, so there is no friction acting on the cannon). Therefore, the total momentum of the two objects (cannon+projectile) must be equal before and after the firing:
But the initial momentum is zero, because at the beginning both the cannon and the projectile are at rest:
So the final momentum, which is sum of the momentum of the cannon and of the projectile, must also be zero:
What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.
<h3>Electrostatics</h3>
I have attached the image of the rod.
We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.
This means that their fields will cancel.
Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.
This also applies to a strong conducting rod and therefore it is strongly attracted.
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Answer:
The velocity of Mosquito with respect to earth will be 0.302m/s
Explanation:
V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air
V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.
Velocity of Mosquito with respect to earth will be
V(me) = V(ma) + V(ae)
We need to find the mosquito’s speed with respect to Earth in the x direction.
V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )
Angle (ae) = –90.0° − 35°=−125°
V(x, me) = 1.10 + (1.4)Cos(-125)
= 1.10 + 1.4(-0.57)
= 1.10 -0.798
= 0.302
So the velocity of Mosquito with respect to earth will be 0.302m/s
We're told that the planets have EQUAL MASS.
If that's true, then the strength of the gravitational forces between
each planet and the star depends only on the distance between
them ... the farther a planet is from the star, the smaller the
gravitational forces are IF we're talking about planets with
equal masses.
Planet-X is closer to the star, and Planet-Y is farther from it.
From this we know that the gravitational forces between the
star and Planet-X are greater, and the forces between the star
and Planet-Y are smaller.
'A' says this.
'B' is totally absurd, because it talks about gravity repelling things.
'C' says exactly the opposite for the two planets.
'D' says that distance doesn't matter. We know this is absurd,
simply because we're never pulled toward Jupiter in our daily life.
