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Valentin [98]
2 years ago
10

What are the limitations of sending information using electronic waves

Physics
1 answer:
vova2212 [387]2 years ago
7 0

Answer:

The limitations of sending information using electromagnetic waves is that when the electromagnetic waves move outward in all directions, wave transmitters need to be focused to transmit their signals to a single specified location.

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An interference pattern is produced by light with a wavelength 580 nm from a distant source incident on two identical parallel s
irakobra [83]

Answer:

Explanation:

1 )

Here

wave length used that is λ = 580 nm

=580 x 10⁻⁹

distance between slit d = .46 mm

= .46 x 10⁻³

Angular position of first order interference maxima

= λ / d radian

= 580 x 10⁻⁹ / .46 x 10⁻³

= 0.126 x 10⁻² radian

2 )

Angular position of second order interference maxima

2 x  0.126 x 10⁻² radian

= 0.252 x 10⁻² radian

3 )

For intensity distribution the formula is

I = I₀ cos²δ/2 ( δ is phase difference of two lights.

For angular position of θ1

δ = .126 x 10⁻² radian

I = I₀ cos².126x 10⁻²/2

= I₀ X .998

For angular position of θ2

I = I₀ cos².126x2x 10⁻²/2

=  I₀ cos².126x 10⁻²

8 0
3 years ago
15 points and brainliest to whoever helps me!
Juliette [100K]

Answer:

The correct answer is false, a plant dying after being exposed to poison is not a physical change nor is it a physical property.

Explanation:

6 0
3 years ago
A water rocket uses an amount of water and pressurized air to send a plastic rocket several feet into the air. As the water and
Ilya [14]

Answer:

D.

For every action there is an equal and opposite reaction.

Explanation:

im doing the same one lol

6 0
2 years ago
Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of t
Lelu [443]

Answer:

c) At a distance greater than r

Explanation:

If G= Gravitational constant

M= Mass of earth

r= distance from earth center

then orbital speed is ;

v = \sqrt{\frac{GM}{r} }

==> v²=GM/r

If speed of first satellite = V₁

==> V₁² = GM/r

==> r = GM/V₁²

If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.

So our answer is c

5 0
2 years ago
Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the
anygoal [31]

Answer:

The value is the temperature of the air inside the tire T_{2} = 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet P_{1} = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet P_{2} = 3.2 + 1 = 4.2 atm

Temperature at inlet T_{1} = 300 K

(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }

\frac{T_{2} }{300}  = \frac{4.2}{3.7}

T_{2} = 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

m_{1}  T_{1} =  m_{2}  T_{2}

\frac{m_{2} }{m_{1}  } = \frac{T_{1} }{T_{2} }

\frac{m_{2} }{m_{1}  } = \frac{300}{354.54}

\frac{m_{2} }{m_{1}  } =0.00294

value of  the original mass of air in the tire should be released is  \frac{m_{2} - m_{1}}{m_{1}}

\frac{0.00294-1}{1}

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

8 0
2 years ago
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