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solong [7]
4 years ago
13

Aless-intense wave will have fewer than a more-intense wave.

Physics
1 answer:
puteri [66]4 years ago
6 0

Decibels

Explanation:

A less intense wave will have fewer decibels than a more intense wave.

The decibel is the unit of measuring the intensity of sound waves.

The intensity of sound wave is the power carried by sound waves in a unit area.

  • Loudness is intimately related to decibels.
  • Loudness is the intensity of sound in the hearing range.

learn more:

Amplitude of sound wave brainly.com/question/2845448

#learnwithBrainly

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A window washer drops a brush from a scaffold on a tall office building. What is the speed of the falling brush after 3.28 s? (N
valina [46]

Answer:

The speed of washer will be 32.144 m/s

Explanation:

Given that

time t= 3.28 s

Acceleration due to gravity

 g=9.8\ \frac{m}{s^2}

Here acceleration  is constant so we cam apply the motion equation

we know that

v= u + at

At initial condition  u =0 m/s

v= u + at

v= 0 + 9.8 x 3.28

v= 32.144 m/s

So the speed of washer will be 32.144 m/s

6 0
3 years ago
Read 2 more answers
Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
larisa86 [58]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • Mass of the larger disk = M_2\ =\ 1.60\ kg
  • Mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the larger disk = R_2\ =\ 5.00\ cm\ =\ 0.05\ m
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)

From the pulley,

\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)

From the equation (1) and (2),

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = R_2\ =\ 0.05\ m

Let a_2 be the acceleration of the block while connecting to the larger pulley.\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2

4 0
3 years ago
N
9966 [12]

Answer: de?

Explanation:

7 0
3 years ago
The ratio of the wavelength of AM radio waves traveling in a vacuum to the wavelength of FM radio waves traveling in a vacuum is
Debora [2.8K]

This question is a big fat non sequitur !

The wavelength of radio waves traveling through vacuum only depends on the frequency that the radio station is licensed to broadcast on, (which had better be the frequency of the transmitter that they buy and use, or they're in big trouble).

The wavelength does NOT depend on the type of modulation that's used to put information onto the signal.  

An amateur radio (ham) operator may very well start out using FM to talk over his radio to somebody else, and then for some reason they may decide to switch to AM.  They can do that without ANY change in the wavelength of their transmissions.

Now, in the USA and many other countries, it so happens that all AM stations are licensed by their governments to transmit their programs on a channel somewhere between 500 KHz and 1.6 MHz, and all FM stations are licensed by their governments to transmit their programs on a channel somewhere between 88 MHz and 108 MHz.  (And THAT's what the radio receivers in these countries are built to receive.)

Then we might say that all of the AM stations are grouped around 1 MHz, and all of the FM stations are grouped around 100 MHz.  The FM frequencies are very roughly 100 times the AM frequencies, so the AM wavelengths are very roughly  100 times the FM wavelengths.  That's <em>choice (3)</em> .

But please don't get the idea that it has anything to do with using AM or FM technology.  It's just a matter of where in the spectrum the government decided to put the AM stations and where they put the FM stations.

For that matter . . . An analog TV station uses an AM signal for the picture and an FM signal for the sound, and it all goes in the same channel, with just about the same wavelengths !

3 0
3 years ago
What is the centripetal acceleration of a small laboratory centrifuge in which the tip of the test tube is moving at 19.0 meters
Volgvan
A_central = v^2/r = (19)^2/10 = 36.1 m/s^2
7 0
3 years ago
Read 2 more answers
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