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3 years ago

Which of the following devices is associated with the reception of radio signals?

Physics

2 answers:

Slav-nsk [51]3 years ago

7 0

Answer: It’s MICROPHONE

Explanation: from PF book

A transmission system at a radio station consists of these devices:

1. Anoscillator,whichconvertsadirectcurrentintoahigh- frequency alternating current

2. Atransmittingantenna,whichsendsoutradiowaveshavingthe same frequency as the current supplied by the oscillator

3. Amicrophone,whichpicksupthesoundwavesandconverts them into a suitable electric current with the same pattern as the sound waves

4. A modulator, which enables the waves produced by the antenna to “carry” the pattern of the sound waves picked up by the microphone

ki77a [65]3 years ago

6 0

What are the "following" devices ? ?
I think they're a list of choices that you have but aren't sharing.

A few devices associated with the reception of various types of
radio signals include the resonant tank, the local oscillator, the
mixer, the detector, the coherer, the discriminator, the parabolic
reflector, the lecher wires, the audio transducer, the demultiplexer,
and ... my personal guess ... the 'antenna' or 'aerial'.

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Freezing Point Depression: Can someone explain this formula to me? ΔTf = Kfcm

Leya [2.2K]

If the solution is treated as an ideal solution, the extent of freezing point depression depends only on the solute concentration that can be estimated by a simple linear relationship with the cryoscopic constant: ΔTF = KF · m · i ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF (solution). KF, the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. Note: When conducting experiments, a higher KF value makes it easier to observe larger drops in the freezing point. For water, KF = 1.853 K·kg/mol.[1] m is the molality (mol solute per kg of solvent) i is the van 't Hoff factor (number of solute particles per mol, e.g. i = 2 for NaCl).

8 0

3 years ago

Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is

attashe74 [19]

Answer:

0.125 volts

Explanation:

The induced emf can be sufficient to stimulate neuronal activity.

One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms.

We need to find the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field. The formula for the induced emf is given by :

$\epsilon=-\dfrac{d\phi}{dt}$

Where

$\phi$ is magnetic flux

So,

$\epsilon=-\dfrac{d(BA)}{dt}\\\\=2\pi r\times \dfrac{dB}{dt}\\\\=2\pi \times 1.6\times 10^{-3}\times \dfrac{1.5-0}{120\times 10^{-3}}\\\\=0.125\ V$

So, the induced emf is equal to 0.125 volts.

7 0

3 years ago

Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20Â°C. Container A is rigid. Container B has a 100 cm^2 pis

Vikki [24]

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

Final Temperature of the gas in A will b GREATER than the temperature in B

Diagram of both processes on a single PV has been uploaded below,

Initial pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) / 0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial pressures in containers A and B is 3039.87 J/liters

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

7 0

3 years ago

For each question, select the right answer from the choices below:

Leni [432]

Option (ii) B is the correct option. The object on the moon has greater mass.

To resolve this, utilize the formulas Force = Mass * Acceleration.

The equation can be used to find the mass given the force in Newtons, using 9.8 m/s² for the acceleration of gravity of the earth and 1.6 m/s² for the moon.

Calculating the mass on earth:

30 N = 9.8 m/s² * mass

This results in a mass of 3.0 kg for the object on Earth.

Calculating the mass of the moon:

30 N = 1.6 m/s²2 * mass

Thus, the moon's object has a mass of 19. kg.

This can be explained by the fact that the earth has a stronger gravitational pull than the moon, producing more force per kilogram of mass. As a result, the moon's mass must be bigger to produce the same amount of force at a lower acceleration from gravity (1.6 m/s² vs. 9.8 m/s²).

To know more about Mass, refer to this link :

brainly.com/question/13386792

#SPJ9

3 0

1 year ago

What safety measures would you suggest to protect from harmful waves

AfilCa [17]

Answer:

If you're talking about the sun than:

Time, Distance and Shielding Time, distance, and shielding actions minimize your exposure to radiation in much the same way as they would to protect you against overexposure to the sun:

If you're talking about the ocean than:

Water safety precautions for teens and young adults:

Never go into the water if you can’t swim.

If you can’t swim, learn. Any age can receive swimming lessons.

Always wear a life jacket while boating or taking part in boating activities such as tubing or skiing.

Never swim alone or in an unsupervised area.

Know your swimming strength.

Don’t rough house around water. Never push, jump on or hang on to others in or around water.

Never drink alcohol while taking part in water or boating activities. Alcohol affects your motor skills therefore making it harder to swim, float, keep balance or drive.

Explanation:

4 0

3 years ago