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Eduardwww [97]
4 years ago
14

For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. What is the

very first frequency of audible sound for these speakers to produce destructive interference at the listener?The speed of sound is 344 m/sec. (1 point)
A. 166 Hz
B. 331 Hz
C. 662 H2
D. 1323 H
Physics
1 answer:
galina1969 [7]4 years ago
7 0

Answer:

The first frequency of audible sound in the speed sound is

f = 662 Hz

Explanation:

vs = 344 m/s

x =  52 cm * 1 / 100m = 0.52m

The wave length is the distance between the peak and peak so

d = 2x

d = 2*0.52 m

d = 1.04 m

So the frequency in the speed velocity is

f = 1 / T

f = vs / x = 344 m/s / 0.52m

f ≅ 662 Hz

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Suppose there was a star with a parallax angle of 1 arcsecond. How far away would it be? Select all that apply.
lesya692 [45]

Answer:

option E

Explanation:

given,                          

Parallax angle(d) = 1 arcsecond

using Parallax formula                  

      d = \dfrac{1}{p}

 p is the parsecs angle which is measured in 1 arcsecond

 d is the distance in parsec

now,                                            

      P = \dfrac{1}{d}

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    1 parsec = 3.26 light year

hence, the answer will be option E

6 0
4 years ago
A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point w
garik1379 [7]

Answer:

The oscillation frequency of the spring is 1.66 Hz.

Explanation:

Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :

E_s=\dfrac{1}{2}kx^2

k is the spring constant

x is the distance to the lowest point form the initial position.

When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.

E_P=mgx

Equating these two energies,

\dfrac{1}{2}kx^2=mgx

\dfrac{k}{m}=\dfrac{2g}{x}.............(1)

The expression for the oscillation frequency is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}} (from equation (1))

f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}

f = 1.66 Hz

So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.

8 0
4 years ago
A 20 N unbalanced force causes an object to accelerate at 1.5 m/s2. What is the mass of the object?
kicyunya [14]

As per Newton's II law we know that

F = ma

here we know that

F = 20 N

a = 1.5 m/s^2

now the mass of the object will be given as

m = \frac{F}{a}

m = \frac{20}{1.5}

m = 13.3 kg

so mass of the object will be 13.3 kg

4 0
3 years ago
A kangaroo hops at an angle of 25° to the horizontal with a velocity of 20 m/s. What is the vertical component of the velocity?
sp2606 [1]

Answer:

A. 8.5 m/s

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vᵧ = 8.5 m/s

6 0
3 years ago
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