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Furkat [3]
3 years ago
6

a new car is advertised as having anti-noise technology. The manufacturer claims that inside the car any sound on negated. Evalu

ate the possibility of such a claim. What would have to be created to cause destructive interference with any sound in the car? Do you believe the manufacturers claim is correct?
Physics
1 answer:
nirvana33 [79]3 years ago
5 0

Answer:

I guess you mean that any sound coming from outside is negated, first, this would mean that you can not hear some signals that may help you to avoid accidents, so you do not want this technology.

Now, let's see if it is possible.

Suppose you have a soundwave approaching the car, the car needs to cancel the soundwaves as the soundwaves approach to the surface, to do it, you need to create another wave that is equal in amplitude, but with a change of phase of pi.

in order to do this, the car must be able to "analyze" the coming soundwave and instantly create another one to cancel it. Suppose that the sound is canceled. now if the sound changes, the car must be able to also change the sound that the car is producing, this instantaneous change is one of the problems.

Another problem is that, as you know, the sound propagates as spherical waves, this means that the wavefront of a wave produced far away will have a bigger radius than the soundwave produced by the car, then we never will have a situation where the wavefront is canceled: This means that we only can cancellate the sound in some areas of the car, and we still will have some sound in other parts of the car.

Another way to isolate the car is with isolating panels, those panels absorb the sound and transmit a very little amount of it (those panels are used in recording studios, for example). With enough of those you can cancel almost all the sound coming from the outside, but to do this you will need a big car because the amount of isolating material needed is a lot.  

We can conclude that the manufacturers claim is not correct.

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A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl
frosja888 [35]

Answer:

W ≅ 292.97 J

Explanation:

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;

W = (Fcosθ)d

Given that:

Tension of the force = 62 N

angle of incline θ =  34°

distance d =5.7 m.

Then;

W = 62 × cos(34) × 5.7

W = 353.4 cos(34)

W = 353.4 × 0.8290

W = 292.9686 J

W ≅ 292.97 J

Hence,  the work done by tension before the block goes up the incline = 292.97 J

8 0
3 years ago
The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.
kiruha [24]

Answer:

3.1\cdot10^{23}\:\mathrm{kg}

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

g_P=G\frac{m}{r^2}., where g_P is acceleration due to gravity at the planet's surface, G is gravitational constant 6.67\cdot 10^{-11}, m is the mass of the planet, and r is the radius of the planet.

Since acceleration due to gravity is given as m/s^2, our radius should be meters. Therefore, convert 2400 kilometers to meters:

2400\:\mathrm{km}=2,400,000\:\mathrm{m}.

Now plugging in our values, we get:

3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2},

Solving for m:

m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}.

6 0
2 years ago
Ultraviolet radiation is dangerous because it has a high enough energy to damage skin cells. What is the BEST explanation of the
MrRissso [65]

The answer is shorter the wavelength, higher the frequency and higher the energy.

<u>Explanation:</u>

The sun radiates UV energy in a wide range of wavelength, which are invisible to human eyes. The shorter the wavelength, the more energetic the radiation, and the greater the potential for harm.

The relationship between wavelength and wave energy is shorter the wavelength, higher the frequency and higher the energy.

It has a frequency ranges from 8 × 10^14 to 3 × 10^16 cycles per second, or hertz (Hz), and wavelengths id about 380 nanometers to about 10 nm.

5 0
4 years ago
The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.0
Alexeev081 [22]

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0  

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If,  V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t  

a(0) = 1.5m/s^²  

1.5m/s^²  =  A + 2B ×  0  

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²  

1.5× 1 +B× 1 = 2m/s

B = 2-1.5  

B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² ×  1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s  

Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

7 0
3 years ago
A rocket is moving away from the solar system at a speed of 7.5 ✕ 103 m/s. It fires its engine, which ejects exhaust with a spee
Elodia [21]

Answer:

a.2.5x 10^3 m/s

b.mr=48kg/s

Explanation:

A rocket is moving away from the solar system at a speed of 7.5 ✕ 103 m/s. It fires its engine, which ejects exhaust with a speed of 5.0 ✕ 103 m/s relative to the rocket. The mass of the rocket at this time is 6.0 ✕ 104 kg, and its acceleration is 4.0 m/s2. What is the velocity of the exhaust relative to the solar system? (B) At what rate was the exhaust ejected during the firing?

velocity of the exhaust relative to the solar system

velocity of the rocket -velocity of the exhaust relative to the rocket.

7.5 ✕ 103 m/s-5.0 ✕ 103 m/s

2.5x 10^3 m/s

. b  we will look for the thrust of the rocket

T=ma

T=6.0 ✕ 104 kg*4.0 m/s2

T=2.4*10^5N

f=mass rate *velocity of the exhaust

T=2.4*10^5N=mr*5.0 ✕ 10^3 m/s

mr=2.4*10^5N/5.0 ✕ 10^3

mr=48kg/s

5 0
3 years ago
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