answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
oksano4ka [1.4K]
2 years ago
7

You are standing at the midpoint between two speakers, a distance D away from each. The speakers are playing the exact same soun

d of frequency 170 Hz in phase with each other. Describe what happens when you walk 1.5 m directly toward one of the speakers. Assume the speed of sound is 340 m/s. Also assume you can't hear any sounds except those produced by the speakers.
A. You begin by hearing a loud sound that gets quieter until you can't hear it anymore. Then it gets louder again until it is as loud as where you started. Then it gets quieter again until you can't hear it anymore.
B. You begin by hearing nothing, but the sound gets louder.
C. You begin by hearing nothing, but the sound gets louder. Then it gets quieter again until you can't hear it anymore.
D. You begin by hearing a loud sound that gets quieter until you can't hear it anymore.
E.You begin by hearing a loud sound that gets quieter until you can't hear it anymore. Then it gets louder again until it is as loud as where you started.
F. You begin by hearing nothing, but the sound gets louder. Then it gets quieter again until you can't hear it anymore. Then it gets louder again.
Physics
1 answer:
Rzqust [24]2 years ago
3 0

Answer:

Explanation:

wave length of sound waves = velocity / frequency

= 340 / 170

λ = 2 m.

When the position of man is exactly at the meddle point between the speakers , sound waves from the speakers reaching man are in same phases ( path difference is zero. ) so intensity of sound is maximum .

Now , the man starts moving towards one of the speakers , his distance from one speaker becomes closer than the other creating path difference for the sound waves reaching his ears.

If he walks a distance of .5 m towards one speaker , path difference created

= .5 x 2 = 1 m

So , path difference = λ /2 ,

there will be destructive interference so minimum sound will be heard there.

When he walks a distance of 1 m , path difference created = 2m

path difference =  λ

so there is constructive interference and maximum  sound will be heard there.

Again we he walks a distance of 1.5 m , path difference created = 3 m

path difference = 3 λ /2

So there will be destructive interference so minimum sound will be heard there.

In this way we see that man starts  from a point of maximum sound intensity , reaches a point of minimum sound intensity , then reaches a point of maximum sound intensity . At last he reaches a position of minimum sound intensity.

You might be interested in
A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi
RSB [31]

Answer:

r = 4.44 m

Explanation:

 

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

         B = ρ g V

Now let's use Newton's equilibrium relationship

         B - W = 0

         B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

         σ = W / A

         W = σ A

The area of ​​a sphere is

           A = 4π r²

       W = W₁ + σ 4π r²

The volume of a sphere is

           V = 4/3 π r³

Let's replace

     ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

     P V = n RT

    P = ρ RT

    ρ = P / RT

 

    P / RT g 4/3 π r³ - σ 4 π r² = W₁

    r² 4π (P/3RT  r - σ) = W₁

Let's replace the values

     r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

     r² (11.81 r -0.060) = 13000 / 4pi

     r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

       r = 4.44 m

3 0
2 years ago
Plastic foam is about 0.10 times as dense as water. What weight of bricks could you stack on a 1m x 1m x 0.10m slab of foam, so
goblinko [34]

Answer: Weight = 98.1N

Explanation:

Density of water = 1000 kg/m^3

Given that the Plastic foam is about 0.10 times as dense as water. That is,

Density of plastic foam = 0.1 × 1000 = 100kg/m^3

The volume V = 1 ×1×0.1 = 0.1 m^3

Density is the ratio of mass to volume

Density = mass/volume

Let us substitute for density and volume to get mass.

100 = M/0.1

Make M the subject of formula

M = 100 × 0.1 = 10 kg

Weight = mg

Where g = 9.81 m/s

Substitute the M and g into the formula

Weight = 10 × 9.81 = 98.1 N

Therefore, the weight of the brick is 98.1 N

4 0
2 years ago
A future use of space stations may be to provide hospitals for severely burned persons. it is very painful for a badly burned pe
natta225 [31]
<span>1.5 minutes per rotation. The formula for centripetal force is A = v^2/r where A = acceleration v = velocity r = radius So let's substitute the known values and solve for v. So F = v^2/r 0.98 m/s^2 = v^2/200 m 196 m^2/s^2 = v^2 14 m/s = v So we need a velocity of 14 m/s. Let's calculate how fast the station needs to spin. Its circumference is 2*pi*r, so C = 2 * 3.14159 * 200 m C = 1256.636 m And we need a velocity of 14 m/s, so 1256.636 m / 14 m/s = 89.75971429 s Rounding to 2 significant digits gives us a rotational period of 90 seconds, or 1.5 minutes.</span>
5 0
2 years ago
Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An
Svetradugi [14.3K]


efficiency= [useful energy transferred ÷ total energy supply]×100%

So, [5500÷10000]×100%=0.55×100

                                        =55%

5 0
2 years ago
An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t
gtnhenbr [62]

1. tan \theta = \frac{v^2}{Rg}

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

N cos \theta = mg (1)

N sin \theta = m \frac{v^2}{R} (2)

where in the second equation, the term m\frac{v^2}{R} represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the  following expression:

tan \theta = \frac{v^2}{Rg}

2. F=\frac{m}{R}(w^2-v^2)

In this second situation, the cars moves around the track at a speed

w>v

This means that the centripetal force term

m\frac{v^2}{R}

is now larger than before, and therefore, the horizontal component of the normal reaction, N sin \theta, is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

N sin \theta + F = m\frac{w^2}{R}

And re-arranging for F,

F=m\frac{w^2}{R}-N sin \theta (3)

But from eq.(2) in the previous part we know that

N sin \theta = m \frac{v^2}{R}

So, susbtituting into eq.(3),

F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)

4 0
2 years ago
Other questions:
  • A sky diver steps from a high-flying helicopter. if there were not air resistance, how fast would she be falling at the end of a
    12·1 answer
  • Sir Marvin decided to improve the destructive power of his cannon by increasing the size of his cannonballs. Sir Seymour kept hi
    13·1 answer
  • A crane with output power of 200W will lift a 600N object a vertical distance of 4.0 meters in seconds
    10·1 answer
  • A 12,000 kg railroad car is traveling at 2 m/s when it strikes another 10,000 kg railroad car that is at rest. If the cars lock
    9·1 answer
  • A car with a mass of 1400kg is being driven along the motorway at 30m/s. Calculate the kinetic energy of the car
    14·1 answer
  • A wood salvage company is hoisting an old tree trunk off the bottom and out of a lake. The cable from the hoist is tied around t
    15·1 answer
  • Rank in order, from largest to smallest, the magnitudes of the horizontal forces F1, F2, and F3 acting on the arrows. Some may b
    12·1 answer
  • Consider two slides, both of the same height. One is long and the other is short. From which slide will a child have a greater f
    12·1 answer
  • A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
    13·1 answer
  • A rock is thrown down from the top of a cliff with a velocity of 3.61 m/s (down). The cliff is 28.4 m above the ground. Determin
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!