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Mumz [18]
3 years ago
8

Objects: A balloon full of helium in the air. Air- 1.27 kg/ml or Helium Balloon- 0.33 kg/ml

Physics
1 answer:
dem82 [27]3 years ago
6 0

Answer:

Air has greater density.

The object will float.

Explanation:

Density of a body is defined as the mass of a body per unit volume of it.

Here, air has a density of 1.27 kg per ml and Helium balloon has a density of 0.33 kg per ml. This means that if volume is 1 ml, then the mass of air in that volume will be 1.27 kg and mass of Helium balloon in that same volume will be 0.33 kg.

Therefore, the mass of Helium balloon is lighter than that of air in a given volume. Hence, air is heavier and has a greater density.

A lighter density object always floats over a greater density body.

Hence, Helium balloon will float over air as Helium balloon is less dense than that of air.

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Explain in your own words the interaction between the electric and magnetic fields that make up a light wave.
iVinArrow [24]

Answer:

They oscillates perpendicularly to one another, the oscillation of one field generates the other field.

Explanation:

In a light wave, an oscillating electric field of a light wave produces a magnetic field, and the magnetic field also oscillates to produce an electric field. The magnetic field and the electric field of a light wave both oscillates perpendicularly to one another. The resultant energy and direction of the wave generated as a result of these oscillating fields is propagated perpendicularly to both fields.

8 0
3 years ago
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

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Answer:

a

Explanation:

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andreyandreev [35.5K]

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