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saw5 [17]
2 years ago
12

Because the sensitivity of the human ear varies over the audio spectrum, how do many sound systems handle very high frequencies

and very low frequencies, respectively, to compensate?
1. dampen; amplify
2. amplify; amplify
3. Sound systems do nothing to compensate.
4. amplify; do nothing to
5. dampen; do nothing to
6. amplify; dampen
7. do nothing to; amplify
8. dampen; dampen
9. do nothing to; dampen
Physics
1 answer:
antoniya [11.8K]2 years ago
3 0

Answer:

Option 2.

Amplify; Amplify

Explanation:

Amplification of sound simply means increasing the loudness of the sound by modifying its electro-acoustic signals. To make very high and very low frequencies of sound audible by the human ear,  the sounds need to be amplified by the sound system amplifier.

Sound systems modify high frequencies by amplifying them, and also, they also modify low frequencies by amplifying them. This changes their wave form and brings them into the hearing range of humans.

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Waves that move the particles of the medium parallel to the direction in which the waves are traveling are called
Nikolay [14]

1. a. longitudinal waves.

There are two types of waves:

- Transverse waves: in transverse waves, the oscillations of the wave occur in a direction perpendicular to the direction of propagation of the wave

- Longitudinal waves: in longitudinal waves, the oscillations of the waves occur parallel to the direction in which the waves are travelling.

So, these types of waves are called longitudinal waves.


2. d. a medium

There are two types of waves:

- Electromagnetic waves: these waves are produced by the oscillations of electric and magnetic field, and they can travel both in a medium and also in a vacuum (they do not need a medium to propagate)

- Mechanical waves: these waves are produced by the oscillations of the particles in a medium, so they need a medium to propagate - therefore, the correct choice is d. a medium


3. a. AM/FM radio

Analogue signals consist of continuous signals, which vary in a continuous range of values. On the contrary, digital signals consist of discrete signals, which can assume only some discrete values. For AM and FM radios, signals are transmitted by using analogue signals.

5 0
2 years ago
What is the net force required to give an automobile with a mass of 1,600 kg an acceleration of 4.5 m/s2
Paladinen [302]

Answer:

Fnet=7200 N

Explanation:

Fnet=mass x acceleration

mass= 1600kg acceleration=4.5m/s^2

8 0
2 years ago
Imagine you derive the following expression by analyzing the physics of a particular system: a=gsinθ−μkgcosθ, where g=9.80meter/
BARSIC [14]

Answer:

Explanation:

Analysis of structure gives

a=gsinθ−μkgcosθ

Notice that all the expression are right but we want to know of we can simplify the expression further.

We want to analyse if we can still further simplify the expression,

Inspecting the Right hand side of the equation, we notice that the acceleration due to gravity is common to both side, so we can bring it out i.e.

So option a is wrong because the expression can be simplified further to

a=g(sinθ−μkcosθ)

Option b is right and the best option.

Since we are given that, g=9.8m/s²

We can as well substitute that to option a

So we will have

a=9.8metre/second²(sinθ−μkcosθ)

Also option C is correct but it is not best inserting the values of g directly without simplifying the expression first

So it will have been the best option if it was written as

a=9.8metre/second²(sinθ−μkcosθ)

So the best option is B.

8 0
2 years ago
A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-s
laiz [17]

Answer:

30298514.82 m/s

Explanation:

M = Mass of star = 2×10³ kg

r = Radius of star = 5×10³ m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

a=G\frac{M}{r^2}\\\Rightarrow a=6.67\times 10^{-11}\frac{2\times 10^{30}}{5\times 10^3}\\\Rightarrow a=2.7\times 10^{16}\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 2.7\times 10^{16}\times 0.017+0^2}\\\Rightarrow v=30298514.82\ m/s

The object would be moving at a velocity of 30298514.82 m/s

5 0
2 years ago
Read 2 more answers
A motor does a total of 480 joules of work in 5.0 seconds to lift a 12-kilogram block to the top of a ramp. The average power de
stepan [7]
Power may be defined as the rate of doing work or the rate of using energy. <span> It is the amount of energy consumed per unit time. It is calculated as follows:

P = E / t
P = 480 / 5
P = 96 W <-----OPTION 3

Hope this answers the question. Have a nice day.</span>
7 0
2 years ago
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