Ask question

Ask question

All categories

olga nikolaevna [1]

3 years ago

14

Two Velocities in a Traveling Wave? Wave motion is characterized by two velocities: the velocity with which the wave moves in th

e medium (e.g., air or a string) and the velocity of the medium (the air or the string itself).
Consider a transverse wave traveling in a string. The mathematical form of the wave is
y(x,t) = A sin(kx-omega...

1 answer:

Lyrx [107]3 years ago

3 0

For the answer to the question above,
<span>There is nothing in the equations to suggest that the string moves in the x direction so D) v_x(x,t)=0.
</span>
y(x,t) = A sin(kx-omega t)
d{y(x,t)}/d{x} = A k cos(kx - omega t)

You might be interested in

If a force of 50 Newton’s was applied to an object with a mass of 500 grams, what will the objects acceleration be?

lapo4ka [179]

Answer: 100 m/s^2

F=ma

Explanation:

50N = 50 kg*m/s^2

500g = 0.5 kg

F=ma

a = F/m

a = (50 kg*m/s^2)/(0.5 kg)

a = 100 m/s^2

5 0

3 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

3 years ago

How do the amplitudes of a 120 decibel sound and a 100-decibel sound compare?

NemiM [27]

The 120 decibel sound has more amplitude than the 100 decibel sound.

In Physics, the relation between amplitude and intensity is that the intensity of the wave is directly proportional to the square of its amplitude.

4 0

2 years ago

What time will the lunar eclipse happen central time?.

alexandr402 [8]

It was about 9:30 p.m. sorry if the answer is wrong

7 0

2 years ago

Calculate the force of attraction between

vagabundo [1.1K]

Answer:

So the force of attraction between the two objects is 3.3365*10^-6

Explanation:

m1=10kg

m2=50kg

d=10cm=0.1m

G=6.673*10^-11Nm^2kg^2

We have to find the force of attraction between them

F=Gm1m2/d^2

F=6.673*10^-11*10*50/0.1^2

F=3.3365*10^-8/0.01

F=3.3365*10^-6

4 0

3 years ago