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3 years ago

What variable affects the natural frequency of an organ pipe?

Physics

2 answers:

const2013 [10]3 years ago

7 0

Pipe Length...i believe that is the correct answer

Misha Larkins [42]3 years ago

5 0

Answer:

The natural frequency of an organ pipe depends on the pipe length.

Explanation:

An organ pipe is an instrument which is used to produce sound. The organ pipe are of two types i.e. closed pipe and organ pipe.

Mathematically, the frequency of closed organ pipe is given as :

$\nu=\dfrac{nv}{4l}$

For open organ pipe :

$\nu=\dfrac{nv}{2l}$

Where,

n is the number of overtones

v is the speed of sound

l is the length of the pipe.

Hence, the correct option is (a) " pipe length ".

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What type of metal are conducting wires most often made of?<br><br> #16

docker41 [41]

Answer:

here

Explanation:

Copper is commonly used as an effective conductor in household appliances and in electrical equipment in general. Because of its low cost, most wires are copper-plated. You will often find electromagnet cores normally wrapped with copper wire

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2 years ago

If an element has 6 electrons, how many of them will be in the 2nd energy level?

Sergio039 [100]

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3 years ago

Monochromatic light falls on a slit that is 2.30×10−3mm wide. Part A If the angle between the first dark fringes on either side

slava [35]

Answer:

Wavelength, $\lambda=1.28\times 10^{-6}\ m$

Explanation:

Given that,

Width of the slit, $d=2.3\times 10^{-3}\ mm=2.3\times 10^{-6}\ m$

The angle between the first dark fringes on either side of the central maximum is 34 degrees

To find,

The wavelength of the light used.

Solution,

The equation of maxima is given by :

$dsin\theta=n\lambda$

n = 1 here

$\lambda=\dfrac{d\ sin\theta}{n}$

$\lambda={d\ sin\theta}$

$\lambda={2.3\times 10^{-6}\times sin(34)}$

$\lambda=1.28\times 10^{-6}\ m$

So, the wavelength of the light used is $1.28\times 10^{-6}\ m$ . Hence, this is the required solution.

5 0

4 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$