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Lady bird [3.3K]
3 years ago
14

Two loudspeakers in a plane, 5.0m apart, are playing the same frequency. If you stand 14.0m in front of the plane of the speaker

s, centered between them, you hear a sound of maximum intensity. As you walk parallel to the plane of the speakers, staying 14.0m in front of them, you first hear a minimum of sound intensity when you are directly in front of one of the speakers.-What is the frequency of the sound? Assume a sound speed of 340 m/s.
Physics
1 answer:
Gre4nikov [31]3 years ago
3 0

Answer:

f= 196.3 Hz

Explanation:

The path difference between two waves given as

\Delta = \sqrt{14^2+5^2}\ -14

Δ = 0.866 m

We know that to hear minimum intensity the path difference given as

Δ = λ/2

λ = 2Δ

We know that

C= f λ

Now by pitting the values

340 = 2 x 0.866 x f

f= 196.3 Hz

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A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
Allisa [31]

Here in this case we can use work energy theorem

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Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

KE = \frac{7}{10}*150*(0.200)^2(50)^2

KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

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7 0
3 years ago
A sheet of metal is 2mm wide 10cm tall and 15cm long. it was 4g. what is the density? <br>​
Hoochie [10]

Answer:

Ro = 133 [kg/m³]

Explanation:

In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

Ro = m/V

where:

m = mass [kg]

V = volume [m³]

We will convert the units of length to meters and the mass to kilograms.

L = 15 [cm] = 0.15 [m]

t = 2 [mm] = 0.002 [m]

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Now we can find the volume.

V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]

And the mass m = 4 [gramm] = 0.004 [kg]

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3 0
2 years ago
g We saw in class that in a pendulum the string does no work. We also saw that the normal force does no work on an object slidin
Ymorist [56]

Answer:

From question (a) and (b) the pendulum motion is perpendicular to the force so the normal force will do no work and the tension in the string of the pendulum will not work

       i.e Normal \ Force(N)  = mg cos \theta

And \theta = 90 so

           N = 0

c

An example will be a where a stone is attached to the end of a string and is made to move in a circular motion while keeping the other end of the string in a fixed position        

d

A dog walking along a surface which has friction, here the frictional force would acting in the direction of the motion and this would do positive work  

Explanation:

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abruzzese [7]

Answer:

The answer is "\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}"

Explanation:

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Luden [163]
The first thing you should know for this case is the definition of distance.
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 answer:
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