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3 years ago

Different between AM and FM

Physics

2 answers:

Hatshy [7]3 years ago

7 0

AM is morning and fmis evening or night till 12

lidiya [134]3 years ago

5 0

The difference is in how the carrier wave is modulated, or altered. With AM radio, the amplitude, or overall strength, of the signal is varied to incorporate the sound information. With FM, the frequency (the number of times each second that the current changes direction) of the carrier signal is varied.

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Before the positive psychology movement, psychology focused mainly on

KonstantinChe [14]

Before positive psychology, the main focus was mostly pathology, that is, studying various psychological issues and sometimes finding ways to treat them, or sometimes not and just studying them for the sake of studying them and noting their occurrence.

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3 years ago

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The mass of a truck with its cargo is 1800 kg. The truck is coasting to the right with a speed of 10 m/s. As it coasts its cargo

OverLord2011 [107]

Here we apply conservation of linear momentum. The momentum of the truck with cargo and without cargo remains constant. That is,

$mv=m'v'$ .

Here $m,v$ are initial mass and velocity. $m',v'$ are final mass and velocity. Here $m=1800,v=10$ and $m'=1800-300=1500$ .

The velocity of the truck be after its cargo is taken off is

$v'=\frac{mv}{m'} \\ v'=\frac{1800*10}{1500} \\ v'=12 m/s$

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4 years ago

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A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

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3 years ago

Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an

Artyom0805 [142]

a) For the motion of car with uniform velocity we have , $s = ut+\frac{1}{2}at^2$ , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.