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2 years ago

4

A loudspeaker at the origin emits a 130 Hz tone on a day when the speed of sound is 340 m/s. The phase difference between two po

ints on the x-axis is 5.6 rad.What is the distance between these two points

1 answer:

Komok [63]2 years ago

4 0

Answer:

distance between two points having phase difference 5.6 rad is

Φ = 2.33 m

Explanation:

given

frequency, f = 130Hz

speed of sound in air, v = 340m/s

distance between two crust or through with phase difference 2 $\pi$ = λ (wavelength)

phase difference = 5.6 rad

note: distance between two points having a phase difference = $\frac{ \lambda }{2\pi }$

∴ distance between two points having phase difference of 5.6 rad is = $\frac{ \lambda }{2\pi }$ ×5.6

Recall

v = f × λ

speed = frequency × wavelength

wavelength = speed/frequency

λ = v/f

λ = 340/130 = 2.615 m

∴ distance between two points having phase difference 5.6 rad is = $\frac{ 2.615 }{2\pi }$ ×5.6 = 2.33 m

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In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc

Vesna [10]

Answer:

Explanation:

Let T be the tension in the swing

At top point $mg-T=\frac{mv^2}{r}$

where v=velocity needed to complete circular path

r=distance between point of rotation to the ball center=L+\frac{d}{2} (d=diameter of ball)

Th-resold velocity is given by $mg-0=\frac{mv^2}{r}$

To get the velocity at bottom conserve energy at Top and bottom

At top $E_T=mg\times 2L+\frac{mv^2}{2}$

Energy at Bottom $E_b=\frac{mv_0^2}{2}$

Comparing two as energy is conserved

$v_0^2=4gr+gr$

$v_0^2=5gr$

$v_0=\sqrt{5gr}$

$v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}$

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2 years ago

What is a limitation of the electron cloud model theory that a law about electrons would not have?

Drupady [299]

Electrons couldn't orbit the nucleus like miniature planets~!

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2 years ago

Read 2 more answers

A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0

2 years ago

What is the difference between 107.6 and 103.26 written with the correct number of significant figures?

andrew-mc [135]

Answer: E. 4.34
To answer this question, you need to understand the principle in subtracting number with a decimal. In this question, the lowest decimal number is from 103.26 which was 2 number. Then you need to convert 107.6 into 107.60 so you can subtract it with 103.26. The answer would be:
107.60 - 103.26= 4.34

4 0

2 years ago

The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is

dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

The angular momentum is defined as:

$L = mrv$ (1)

Since there is no torque acting on the system, it can be expressed in the following way:

$t = \frac{\Delta L}{\Delta t}$

$t \Delta t = \Delta L$

$\Delta L = 0$

$L_{a} - L_{p} = 0$

$L_{a} = L_{p}$ (2)

Replacing equation 1 in equation 2 it is gotten:

$mr_{a}v_{a} =mr_{p}v_{p}$ (3)

Where m is the mass of the comet, $r_{a}$ is the orbital radius at the aphelion, $v_{a}$ is the speed at the aphelion, $r_{p}$ is the orbital radius at the perihelion and $v_{p}$ is the speed at the perihelion.

From equation 3 v_{a} will be isolated:

$v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}$

$v_{a} = \frac{r_{p}v_{p}}{r_{a}}$ (4)

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

$r_{p} = 1.69 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $2.528x10^{11} m$

$r_{p} = 2.528x10^{11} m x \frac{1km}{1000m}$ ⇒ $252800000 km$

$r_{a} = 4.40 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $6.582x10^{11} m$

$r_{p} = 6.582x10^{11} m x \frac{1km}{1000m}$ ⇒ $658200000 km$

Then, finally equation 4 can be used:

$v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}$

$v_{a} = 10.75 km/s$

Hence, the speed at the aphelion is 10.75 km/s.

8 0

2 years ago