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bonufazy [111]
3 years ago
7

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt

ed by each speaker is 632 Hz . You are standing between the speakers, along the line connecting them and are at a point of constructive interference.How far must you walk toward speaker B to move to reach the first point of destructive interference? (Assume that the speed of sound in air is 344 m/s).
Physics
1 answer:
Andrews [41]3 years ago
7 0

<h2>The distance covered by man is 6.4 m</h2>

Explanation:

In the interference of two waves , these will fall upon each other .

It will make nodes and anti-nodes in between . There will be one node in between two anti-nodes and  will be one anti-node between two nodes .

The distance between two nodes is λ/2 and distance between two anti-nodes is also λ/2

Thus if the man is at constructive interference , so to move at destructive interference , it has to cover a distance equal to λ/4

Here λ = velocity/frequency = 344/632 = 18.4 m

Thus the distance he will cover = 18.4/4 = 4.6 m

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Explanation:

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Calculate the time of fight for a horizontally launched projectile from a height of 20m above the ground with an initial velocit
Hoochie [10]

Answer:

0.5sec

Explanation:

Parameters

Height(H) =20m

Initial velocity(u) = 5m/s

Acceleration due to gravity(g) =10m/s^2

Method one

Using the second law of motion

H=ut-1/2gt^2

20=5t-1/2×10×t^2

20=5t-5t^2

dh/dt = 5-10t

where any constant is zero therefore the 20 is zero

5-10t=0

Collect like terms

-10t= -5

t=1/2 = 0.5sec

2nd method

Parameters

Height(H) =20m

Initial velocity(u) = 5m/s

Acceleration due to gravity(g) =10m/s^2

Using the time taken formula

t=u/g

t=5/10

t=0.5sec

4 0
3 years ago
A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied
Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that  the minimum force that must be applied on the <em>book is</em>

  • F=44N

From the question we are told

A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the  Force  is mathematically given as

F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\

F=44N

Therefore

the minimum force that must be applied on the <em>book is</em>

F=44N

For more information on this visit

brainly.com/question/23379286

8 0
3 years ago
Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

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