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
irinina [24]
2 years ago
13

To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d

= 4.00 m apart. A listener observes maximum constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point halfway between the speakers is l = 5.00 m . One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r = 60.0 cm from its original position, the listener, who is not moving, observes destructive interference for the first time. Find the speed of sound v in the air if both speakers emit a tone of frequency 700 Hz .

Physics
1 answer:
Nimfa-mama [501]2 years ago
7 0

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is  v  =  350 \  m/s  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

  The change in the distance of the speaker is by k  =  60 cm  =  0.6 \  m

    The frequency of both speakers is f =  700 \  Hz

Generally the distance of the listener to the first speaker is mathematically represented as

       L_1  =  \sqrt{l^2 + [\frac{d}{2} ]^2}

       L_1  =  \sqrt{5^2 + [\frac{4}{2} ]^2}

        L_1  =   5.39 \  m

Generally the distance of the listener to second speaker at its new position is  

          L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}

       L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}

        L_2  =   5.64  \  m  

Generally the path difference between the speakers is mathematically represented as

        pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}

Here \lambda is the wavelength which is mathematically represented as

         \lambda =  \frac{v}{f}

=>    L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    5.64 - 5.39   =  \frac{1  *  v}{2*700}      

=>    v  =  350 \  m/s  

You might be interested in
Which of the following sketches represents a possible configuration for this problem?
garri49 [273]
Where are the following sketches?
7 0
2 years ago
How much is the moment due to force P about point B. P(unit=N) vector is equal to (150i+260j ) and vector BA (unit=meter) is equ
artcher [175]

Answer:

The moment (torque) is given by the following equation:

\vec{\tau} = \vec{r} \times \vec{F}\\\vec{r} \times \vec{F} = \left[\begin{array}{ccc}\^{i}&\^j&\^k\\r_x&r_y&r_z\\F_x&F_y&F_z\end{array}\right] = \left[\begin{array}{ccc}\^{i}&\^j&\3k\\0.23&0.04&0\\150&260&0\end{array}\right] = \^k((0.23*260) - (0.04*150)) = \^k (53.8~Nm)

Explanation:

The cross-product between the distance and the force can be calculated using the method of determinant. Since the z-components are zero, it is easy to calculate.

4 0
1 year ago
Glider‌ ‌A‌ ‌of‌ ‌mass‌ ‌0.355‌ ‌kg‌ ‌moves‌ ‌along‌ ‌a‌ ‌frictionless‌ ‌air‌ ‌track‌ ‌with‌ ‌a‌ ‌velocity‌ ‌of‌ ‌0.095‌ ‌m/s.‌
NemiM [27]

Answer:

vB' = 0.075[m/s]

Explanation:

We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.

Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.

(m_{A}*v_{A})+(m_{B}*v_{B})=(m_{A}*v_{A'})+(m_{B}*v_{B'})

where:

mA = 0.355 [kg]

vA = 0.095 [m/s] before the collision

mB = 0.710 [kg]

vB = 0.045 [m/s] before the collision

vA' = 0.035 [m/s] after the collision

vB' [m/s] after the collison.

The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.

(0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]

7 0
1 year ago
The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational
s344n2d4d5 [400]

Answer:

v = √2G M_{earth} / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

        Eo = K + U = ½ m1 v² - G m1 m2 / r1

        Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

       Eo = Ef

       ½ m1v² - G m1 M_{earth} / R = - G m1 M_{earth} / R

      v² = 2G M_{earth} (1 / R - 1 / Rinf)

If we do Rinf = infinity     1 / Rinf = 0

       v = √2G M_{earth} / R

      Ef = = - G m1 m2 / R

The mechanical energy is conserved  

 

      Em = -G m1  M_{earth} / R

      Em = - G m1  M_{earth} / R

     R = int        ⇒  Em = 0

6 0
1 year ago
A gold puck has a mass of 12 kg and a velocity of 5i – 4j m/s prior to a collision with a stationary blue puck whose mass is 18
Ugo [173]

Answer:

Explanation:D

6 0
2 years ago
Other questions:
  • What is the speed of each neutron as they crash together? keep in mind that both neutrons are moving?
    9·2 answers
  • A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate
    9·1 answer
  • How many neutrons are contained in 2 kg? Mass of one neutron is 1.67x10-27 kg.
    5·1 answer
  • In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li
    14·1 answer
  • A 2-kg wood block is pulled by a string across a rough horizontal floor. The string exerts a tension force of 30 N on the block
    13·1 answer
  • In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that the gas expands with no add
    15·1 answer
  • An astronaut on a space walk floats a little too far away
    5·2 answers
  • A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. Determine the
    12·1 answer
  • If you're swimming underwater and knock two rocks together, you will hear a very loud noise. But if your friend above the water
    7·1 answer
  • Pamela has three computers, all of
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!