Wavelength multiplied by frequency equals

Speakers A and B are vibrating in phase. They are directly facing each other, are 8.0 m apart, and are each playing a 75.0-Hz to

NeX [460]

Answer:

4 m, 1.71 m and 6.29 m

Explanation:

Let L = 8 m be the distance between the two speakers. Let x be the distance from speaker A of constructive interference. The distance to speaker B from the point of constructive interference is thus x₁ = L - x.

There is constructive interference when the distance x₁ - x = nλ where n = is an integer and λ = wavelength L - x

x₁ - x = nλ

L - x - x = nλ

L - 2x = nλ

x = (L - nλ)/2 = (L - nv/f)/2. where v = speed of wave = 343 m/s and f = frequency = 75 Hz

The distance from A where constructive interference would occur starts from when

n = 0

x₂ = (L - nv/f)/2 = (8 - 0 × 343/75)/2 = (8 - 0)/2 = 8/2 = 4 m

n = 1

x₃ = (L - nv/f)/2 = (8 - 1 × 343/75)/2 = (8 - 4.57)/2 = 3.43/2 = 1.71 m

when n = 2

x₄ = (L - nv/f)/2 = (8 - 2 × 343/75)/2 = (8 - 9.14)/2 = -1.15/2 = -0.57 m

So the value at n = 2 is not included.

The third point occurs at x₅ = L - x₃ where x₃ = 1.71 m is the distance away from point B where constructive interference also occurs. (since it is symmetrical about the point x₂ = 4 m

x₅ = L - x₃ = 8 - 1.71 = 6.29 m

8 0

3 years ago

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Use Hooke's Law to determine the work done by the variable force in the spring problem. Nine joules of work is required to stret

natima [27]

Answer:

29.16 J

Explanation:

From Hook's law,

W = 1/2(ke²)..................... Equation 1

Where W = work done, k = Spring constant, e = extension.

Given: W = 9 J, e = 0.5 m.

Substitute into equation 1

9 = 1/2(k×0.5²)

Solve for k

k = 18/0.5²

k = 72 N/m.

The work done required to stretch the spring by additional 0.4 m is

W = 1/2(72)(0.4+0.5)²

W = 36(0.9²)

W = 29.16 J.

6 0

3 years ago

A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that

saveliy_v [14]

Answer:

The speed it reaches the bottom is

$v=6.51m/s$

Explanation:

Given: $m=5.0kg$ , $r=47cm\frac{1m}{100cm}=0.47m$

Using the conservation of energy theorem

$U_i=K_E+K_{ER}$

$m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2$

$v=r*w$ , $I=\frac{1}{2}*m*r^2$

$m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2$

$m*g*h=\frac{3}{4}*m*r^2*w^2$

$g*h=\frac{3}{4}*r^2*w^2$

Solve to w'

$w^2=\frac{4*g*h}{3*r^2}$

$h=x*sin(30)=6.5m*sin(30)=3.25m$

$w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}$

$w=27.74rad/s$

$v=27.74rad/s*0.235m=6.51m/s$

7 0

3 years ago

Question 20

oksian1 [2.3K]

Answer:

The image distance is 17.56 cm

Explanation:

We have,

Height of light bulb is 3 cm.

The light bulb is placed at a distance of 50 cm. It means object distance is, u =-50 cm

Focal length of the lens, f = +13 cm

Let v is distance between image and the lens. Using lens formula :

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{13}+\dfrac{1}{(-50)}\\\\v=17.56\ cm$

So, the image distance is 17.56 cm.

5 0

3 years ago

Which statement is true regarding flexibility? A. joint's range of motion cannot be maintained with age. b.A joint's range of mo

Alex777 [14]

<span>the statement that is true regarding flexibility is : b. a joint's range of motion will be lost if the joint is not used regularly. Our body is like a machine. If we not constantly heat it up, our body will be more prone to injury. We can see that the old people who lived within the tribe in the middle of the mountain are far stronger than the one who lived in the city.</span>

3 0

3 years ago

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