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3 years ago

How can a wave carry information?

2 answers:

7 0

A transmitter “encodes” or modulates messages by varying the amplitude or frequency of the wave – a bit like Morse code. At the other, a receiver tuned to the same wavelength picks up the signal and 'decodes' it back to the desired form

I think it’s A or D

Katen [24]3 years ago

4 0

Answer:

i think its A

Explanation:

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Light of a given wavelength is used to illuminate the surfacce of a metal, however, no photoelectrons are emitted. In order to c

Jlenok [28]

Answer:

B. use light of a shorter wavelength.

Explanation:

We know that

$E= \frac{hc}{\lambda}$

h= plank's constant

c= speed of light

λ= wavelength of the incident light

so, in order to have sufficient energy for for the emission of electron, the incident's radiation must have λ small enough.

B. use light of a shorter wavelength.

6 0

3 years ago

Gasoline burns in the cylinder of an automobile engine. During the combustion reaction, the production of gas forces the piston

serg [7]

Answer:

$\Delta U = 1640 J$

Explanation:

As we know by first law of thermodynamics that for ideal gas system we have

Heat given = change in internal energy + Work done

so here we will have

Heat given to the system = 2.2 kJ

Q = 2200 J

also we know that work done by the system is given as

$W = 560 J$

so we have

$\Delta U = Q - W$

$\Delta U = 2200 - 560$

$\Delta U = 1640 J$

6 0

3 years ago

What is the desired tone of an instrument called?

Alik [6]

Answer:

Timbre also known as color or tone quantity which is used for music notes sound and tone are mostly know for guitars or pianos .

Does this help .

6 0

3 years ago

Read 2 more answers

4. X-ray radiation from diffraction and fluorescence instruments is a) Not very dangerous because the X-rays are absorbed in the

Dahasolnce [82]

Answer:

c) Very dangerous and users must not override devices designed to protect from exposures.

Explanation:

X-ray is a form of high energy electromagnetic radiation and are part of the electromagnetic spectrum.

X-ray radiation from diffraction and fluorescence instruments is very dangerous because of their high energy and wavelength.

Hence, users must not override devices designed to protect from exposures. The best shielding device to protect one from exposure is Lead.

7 0

3 years ago

A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal

zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity $u_{y}=0$

The initial velocity is the horizontal component $u_{x}$

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = $u_{y}$ t + $\frac{1}{2}$ gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , $u_{y}=0$

Substitute these values in the rule

→ -60 = 0 + $\frac{1}{2}$ (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* The ball is in the air for 3.5 seconds 

The initial velocity is the horizontal component $u_{x}$

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = $u_{x}$ t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = $u_{x}$ (3.5)

Divide both sides by 3.5

→ $u_{x}$ = 28.57 m/s

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is $v_{y}$

→ $v_{y}$ = $u_{y}$ + gt

→ $u_{y}$ = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ $v_{y}$ = 0 + (-9.8)(3.5)

→ $v_{y}$ = -34.3 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity v is the resultant vector of $v_{x}$ and $v_{y}$

→ Its magnetude is $v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}$

→ Its direction $tan^{-1}\frac{v_{y}}{v_{x}}$

→ $v_{y}$ = 28.6 , $v_{y}$ = -34.3

Substitute this values in the rules above

→ $v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66$

→ Its direction $tan^{-1}\frac{-34.3}{28.6}=-50.18$

The negative sign means the direction is below the horizontal

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

7 0

3 years ago