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KonstantinChe [14]

2 years ago

15

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 20 cm apart. The sound

intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 70 cm.
Required:
a. What is the wavelength of the sound?
b. If the distance between the speakers continues to increase, at what separation will the sound intensity again be a maximum?

1 answer:

krek1111 [17]2 years ago

4 0

Answer:

Explanation:

The path length difference = extra distance traveled

The destructive interference condition is:

$\Delta d = (m+1/2)\lambda$

where m =0,1, 2,3........

So, ←

$\Delta d = (m+1/2)\lamb da9/tex]so [tex]\Delta d = \frac{\lambda}{2}$

⇒ λ = 2Δd = 2×10 = 20

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Vector ~A has a magnitude of 29 units and points in the positive y-direction. When vector ~B is added to ~A, the resultant vecto

timurjin [86]

Answer:

The magnitude of vector B is 43 units and it points in the negative y-direction.

Explanation:

Resultant of vectors = vector sum of all the vectors

Vector A = 29j

Vector B = ?

Resultant of vector A and B = R = -14j

R = A + B

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B = -14j - 29j = - 43j

Hence, the magnitude of vector B is 43 units and it points in the negative y-direction.

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3 years ago

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

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3 years ago

Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom

Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

final velocity of Tarzan = v_f

law of conservation of energy

PE_i + KE_i = PE_f + KE_f

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i = \dfrac{1}{2}mv_f^2$

$v_f = \sqrt{2gh_i}$

$= \sqrt{2gL(1- cos\theta)}$

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= 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

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$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2$

$gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2$

$v_f = \sqrt{v_1^2+2gh_i}$

$v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}$

v_f= 11.29 m/s

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A 0.20 kg mass attached to the end of a spring causes it to stretch 3.0 cm. What is the spring constant? What is the potential e

SVEN [57.7K]

Given that the mass is m = 0.2 kg and the displacement is x = 3 cm = 0.03 m

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The spring constant can be calculated by the formula

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Here, k is the spring constant.

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