All categories

3 years ago

Which of the following do waves carry?

Physics

1 answer:

kumpel [21]3 years ago

8 0

I would go with energy, because they give alllooooot of energy, from the waves.
hope that this would help you! =)

You might be interested in

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt

Tamiku [17]

Answer:

7 m .

Explanation:

For destructive interference

Path difference = odd multiple of λ /2

Wave length of sound from each of A and B.

= speed / frequency

λ = 334 / 172 = 2 m

λ/2 = 1 m

If I am 1 m away from B , the path difference will be

8 - 1 = 7 m which is odd multiple of 1 or λ /2

So path difference becomes odd multiple of λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

8 0

3 years ago

Read 2 more answers

Can someone please help meee!!

Paraphin [41]

The answer is 9800 J/S

5 0

2 years ago

2.- Si una cámara fotográfica emite un pulso de sonido para enfocar un objeto, determinar

uranmaximum [27]

Answer:

a. El tiempo de recorrido es $5.882\times 10^{-3}$ segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

Explanation:

El sonido es un tipo de onda mecánica, que es un tipo de onda que necesita de un medio material para propagarse. En este caso, entendemos que el sonido se propaga a través del aire atmosférico hasta llegar a su destino y devolverse a rapidez constante. Entonces, podemos estimar el tiempo ( $t$ ), medido en segundos, a partir de la siguiente fórmula:

$t = \frac{2\cdot x_{s}}{v_{s}}$

Donde:

$x_{s}$ - Distancia entre la cámara fotográfica y el objeto, medida en metros.

$v_{s}$ - Rapidez del sonido en el aire atmosférico, medida en metros por segundo.

A continuación, calculamos el tiempo de recorrido:

a. ( $x_{s} = 1\,m$ , $v_{s} = 340\,\frac{m}{s}$ )

$t = \frac{2\cdot (1\,m)}{340\,\frac{m}{s} }$

$t = 5.882\times 10^{-3}\,s$

El tiempo de recorrido es $5.882\times 10^{-3}$ segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. ( $x_{s} = 20\,m$ , $v_{s} = 340\,\frac{m}{s}$ )

$t = \frac{2\cdot (20\,m)}{340\,\frac{m}{s} }$

$t = 0.118\,s$

El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

8 0

2 years ago

Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in

maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

$v^{2} = \frac{dr}{dt}^{2} +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2}$ - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

$\frac{r}{z}=tan(45)$

or:

$z = r cot(45)$ - (2)

and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

Now the kinetic energy is given as:

$T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2})$ - (4)

And the potential energy is given by:

$V = -mgz = -mgr cot(45)$

So the Langrangian is given by:

$L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)$

And the equations of motion are:

For θ

$\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c$

For r

$\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}$