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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) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
It is known as a basic unit of all living things.
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cell

4 0
2 years ago
A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider
julia-pushkina [17]

There's not enough information to find an answer.

I think the idea here is that in descending (416 - 278) = 138 meters,
the glider gives up some gravitational potential energy, which
becomes kinetic energy at the lower altitude.  This is all well and
good, but we can't calculate the difference in potential energy
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3 0
3 years ago
A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i
galben [10]

Answer:

350 F to 100 F it take approx 87.33 min  

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling  

\frac{dT}{dt} = -k(T-Ta)

\frac{dy}{dt} = \frac{d}{dt} (T(t) -Ta)

= \frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt} = -k(T-Ta)

-ky \frac{dy}{dt} = -ky

T(t) -Ta = (To -Ta) e^{-kt} T(t) = Ta+ (To -Ta)  e^{-kt}

put her value for time 30 min and T(t) = 200◦F and To =350◦F  and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) e^{-k30}

k = 0.025575

so here for  T(t) = 100F

100 = 70 + ( 350 - 70 ) e^{-0.025575*t}

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min  

5 0
3 years ago
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