Two loudspeakers, 4.0 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m.a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.25 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Answers

Answer 1

Answer:

Answer:

a) 343.0 Hz b) 686.0 Hz

Explanation:

a) First, we need to know the distance to both speakers.

If the person is at halfway between the two speakers, and they are 4.0 m apart, this means that he is at 2.0 m from each speaker.

So, if he moves 0.25 m towards one of them, the distance from any speaker will be as follows:

d₁ = 2.0 m-0.25 m= 1.75 m

d₂ = 2.0 m + 0.25 m = 2.25 m

The difference between these distances is the path difference between the sound from both speakers:

d = d₂ - d₁ = 2.25 m - 1.75 m = 0.5 m

If the person encounters at this path difference a minimum of sound intensity, this means that this distance must be an odd multiple of the semi-wavelength:

d = (2*n-1)*(λ/2) = 0.5 m

The minimum distance is for n=1:

⇒ λ = 2* 0.5 m = 1 m

In any wave, there exists a fixed relationship between the speed (in this case the speed of sound), the wavelength and the frequency, as follows:

v = λ*f, where v= 343 m/s and λ=1 m.

Solving for f, we have:

$f =(343.0 m/s)/(1.0 m) = 343 Hz$

b) If the person remains at the same point, for this point be a maximum of sound intensity, now the path difference (that it has not changed) must be equal to an even multiple of the semi-wavelength, which means that it must be met the following condition:

d = 0.5 m = 2n*(λ/2) = λ (for n=1)

if the speed remains the same (343 m/s) we can find the new frequency as follows:

$f =(v)/(d) =(343 m/s)/(0.5m) =686.0 Hz$

⇒ f = 686.0 Hz

Answer 2

Answer:

Final answer:

Two speakers create peaks and troughs of sound intensity due to constructive and destructive interference of waves. Using wave properties, the frequency of the sound when a minimum intensity is experienced 0.25m from the center is 680Hz. Increasing the frequency, the first to produce maximum intensity at the same position is about 2720Hz.

Explanation:

The behavior of sound intensity in this question is due to wave interference, specifically, constructive and destructive interference of sound waves. When you stand halfway between the speakers, the sound waves from each speaker are in phase, which means the pressure variations combine to create an intensified sound, known as constructive interference.

When you move towards one of the speakers and detect a minimum in sound intensity, this is due to destructive interference, which occurs when the crest of one wave overlaps with the trough of another, canceling each other and producing a minimum sound level.

a. The frequency of the sound can be calculated using the formula for wave speed, v = f.lambda, where v is the speed of sound (340 m/s under normal conditions), f is the frequency, and lambda is the wavelength. In this case, a minimum sound intensity indicates one-half wavelength. So, lambda = 0.5 m. Thus, frequency, f = v/lambda = 340/0.5 ~ 680 Hz.

b. When you increase the frequency while remaining 0.25m from the center, the first frequency for which the location will be a maximum of sound intensity will be when you are an integral multiple of the wavelength away from the source. Thus if we let this be 2λ, we can calculate the frequency as f = v / λ = v / (0.25m / 2) = 340 / 0.125 ~ 2720 Hz.

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A businessperson took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours, and each way the trip was 200 miles. What was the speed of the wind that affected the plane, which was flying at a speed of 120mph? Round your answer to the nearest whole number.

Answers

Answer:

Speed of the wind is 48.989 mph

Explanation:

We have given each trip is of 200 miles

So total distance = 200 +200 = 400 miles

Speed of the airplane = 120 mph

Let the speed of the wind = x mph

So the speed of the airplane with wind = 120+x

So time taken by airplane with wind = $(200)/(120+x)$

Speed of the airplane against the wind = 120 - x

So time taken by the airplane against the wind $=(200)/(120-x)$

Total time is given as t= 4 hour

So $(200)/(120+x)+(200)/(120-x)=4$

$(200(120-x)+200(120+x))/((120+x)(120-x))=4$

$48000=57600-4x^2$

$4x^2=9600$

x = 48.989 mph

Answer:

Explanation:

Type Distance Rate Time

Headwind 200 120-r 200/120-r

Tailwind 200 120 - r 200/120 - r

We know the times add to 4, so we write the equation:

200/120−r + 200/120 + r = 4

We multiply both sides by the LCD and simplify to get:

(120−r)(120+r) ((200/120 -r ) + 200/120+r) = 4(120 -r) (120 +r)

200(120−r)+200(120+r)=4(120−r)(120+r)

Factor the 200 and simplify inside the parentheses to find:

200(120−r+120+r)=4(1202−r2)

200(240)=4(1202−r2)

200(60)=120^2−r^2

12,000=14,400−r^2

−2,400= −r^2

49 ≈ r

The speed of the wind is 49mph.

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answers

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_(x) = P+T-\mu_(k)\cdot N = 0$ (Ec. 1)

$\Sigma F_(y) = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_(k)$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_(k)\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_(k) =(P+T)/(m\cdot g)$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then:

$\mu_(k) = (400\,N+290\,N)/((300\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right))$

$\mu_(k) = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

Final answer:

The calculation of the coefficient of kinetic friction involves setting the total force exerted by the workers equal to the force of friction, as the crate moves at a constant speed. The coefficient of kinetic friction is then calculated by dividing the force of friction by the normal force, which is the weight of the crate. The coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Explanation:

To calculate the coefficient of kinetic friction, we first must understand that the crate moves at a constant velocity, indicating that the net force acting on it is zero. Thus, the total force exerted by the workers (400 N + 290 N = 690 N) is equal to the force of friction acting in the opposite direction.

Since the frictional force (F) equals the normal force (N) times the coefficient of kinetic friction (μk), we can write the equation as F = μkN. Here, the normal force is the weight of the crate, determined by multiplying the mass (m) of the crate by gravity (g), i.e., N = mg = 300 kg * 9.8 m/s² = 2940 N.

Next, we rearrange the equation to solve for the coefficient of kinetic friction: μk = F / N. Substituting the known values (F=690 N, N=2940 N), we find: μk = 690 N / 2940 N = 0.2347. Thus, the coefficient of kinetic friction for the crate on the floor is approximately 0.235.

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In the circuit shown in the figure above, suppose that the value of R1 is 100 k ohms and the value of R2 is 470 k ohms. At which of the following locations in the circuit would you measure the highest voltage with your meter? A. Between points A and B
B. Between points A and C
C. Between points B and E
D. Between points B and C

Answers

Answer:

Explanation:

voltage between A and C is equal battery's voltage.

The measurement of an electron's energy requires a time interval of 1.2×10^−8 s . What is the smallest possible uncertainty in the electron's energy?

Answers

Answer:

$1.05* 10^(-26)J$

Explanation:

The uncertainty in energy is given by $\Delta E=(h)/(2\pi \Delta t)$

here h is plank's constant which value is $6.67* 10^(-34)$ and $\Delta t$ is the time interval which is given as $1.2* 10^(-8)sec$

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A train station bell gives off a fundamental tone of 505 Hz as the train approaches the station at a speed of 27.6 m/s. If the speed of sound in air on that day is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train

Answers

Answer:

Apparent frequency of the bell to the observer is 546.12 Hz

Explanation:

The frequency of train bell (frequency of source) = 505 Hz

The speed of train (observer) = 27.6 m/s

The speed of sound in the air is (velocity of sound) = 339 m/s

The apparent frequency of the bell to the observer is calculated as follows:

Apparent frequency of bell to the observer.

$= \text{frequency of source} * (Observer + velocity \ of \ sound )/( velocity \ of \ sound ) \n= 505 * (27.6 + 339)/(339) \n= 546.12 Hz$

Which statement describes how a machine can help make work easier? It can put out more force than the input force by decreasing the distance over which force is applied. It can increase the amount of work performed if the output force is in the same direction as the input force. It can apply a greater output force over a greater distance compared to the input force and distance. It can increase the distance over which output force acts by applying less output force than input force.

Answers

Answer:

It can apply a greater output force over a greater distance compared to the input force and distance.

Explanation:

A machine helps to multiply force input and makes it easier to do work. Simple and complex machines are need to make work easier.

Most machines are designed to increase the input force by a system of mechanics.
Machines allow force to be applied over great distances and this allows for the input work done to be duly compensated for.

The statement describes how a machine can help make work easier, (c) It can apply a greater output force over a greater distance compared to the input force and distance is correct option.

A machine is a device designed to perform a specific task or work by using mechanical, electrical, or other forms of energy. Machines are created to make tasks easier, more efficient, or possible to accomplish that would be difficult or impossible to achieve using human effort alone.

They can range from simple tools like levers and pulleys to complex systems like engines and computers. Machines typically involve the conversion of input energy (such as human effort or electricity) into some form of output energy (such as mechanical motion or computation) to achieve a desired result.

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