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german
3 years ago
13

A closed-end organ pipe is used to produce a mixture of sounds. The third and fifth harmonics in the mixture have frequencies of

1100 Hz and 1833 Hz respectively. What is the frequency of the first harmonic played by the organ pipe
Physics
1 answer:
nexus9112 [7]3 years ago
7 0

Answer:

F_1=366.67Hz

Explanation:

From the question we are told that:

Frequency of 3rd harmonics F_3=1100

Frequency of 5th harmonics F_3=1833

Generally the equation for Wavelength at 3rd Harmonics is mathematically given by

 \lambda_3=\frac{4}{3}l

Therefore

 F_3=\frac{3v}{4l}

Generally the equation for Wavelength at 1st Harmonics is mathematically given by

 \lambda_1=\frac{4}{1}l

Therefore

 F_1=\frac{v}{4l}

Generally the equation for the frequency of the first harmonic is mathematically given by

 F_1=\frac{F_3}{3}

 F_1=\frac{1100}{3}

 F_1=366.67Hz

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Consider two diffraction gratings. One grating has 3000 lines per cm, and the other one has 6000 lines per cm. Both gratings are
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Answer:

<em>The 6000 lines per cm grating, will produces the greater dispersion .</em>

Explanation:

A diffraction grating is an optical component with a periodic (usually one  that has ridges or rulings on their surface rather than dark lines) structure that splits and diffracts light into several beams travelling in different directions.

The directions of the light beam produced from a diffraction grating depend on the spacing of the grating, and also on the wavelength of the light.

For a plane diffraction grating, the angular positions of principle maxima is given by

(a + b) sin ∅n = nλ

where

a+b is the distance between two consecutive slits

n is the order of principal maxima

λ is the wavelength of the light

From the equation, we can see that without sin ∅ exceeding 1, increasing the number of lines per cm will lead to a decrease between the spacing between consecutive slits.

In this case, light of the same wavelength is used. If λ and n is held constant, then we'll see that reducing the distance between two consecutive slits (a + b) will lead to an increase in the angle of dispersion sin ∅. So long as the limit of sin ∅ not greater that one is maintained.

7 0
3 years ago
Objects P and Q are placed in a bowl of water. P floats and Q sinks. Which one or more of the following can be deduced from this
Lady bird [3.3K]

The correct option that can be deduced for both Object P and Q is Option b) I and II only

To solve this question correctly, we need to understand the concept of density and it relation to mass and volume.

<h3>What is Density?</h3>

Density is a physical property of an object and can be expressed by using the relation:

\mathbf{Density = \dfrac{mass}{volume}}

From the given parameters, we are being told that:

  • P → floats
  • Q → sinks

This implies that Q has a greater density that P. Since Q has a greater density than P, Q will be heavier since it will have greater mass.

However, Q will not be denser than water because if that happens, P will be have a greater density which is untrue in this scenario.

Therefore, we can conclude that:

  • 1. Q is heavier than P
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Learn more about density here:

brainly.com/question/6838128

6 0
3 years ago
Kelli weighs 425 N, and she is sitting on a playground swing that hangs 0.36 m above the ground. Her mom pulls the swing back an
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Answer:

V = 3.54 m/s

Explanation:

Using the conservation of energy:

E_i = E_f

so:

wh = \frac{1}{2}mV^2

where w is te weigh of kelly, h the distance that kelly decends, m is the mass of kelly and V the velocity in the lowest position.

So, the mass of kelly is:

m = 425N/9.8 = 43.36 Kg

and h is:

h = 1m-0.36m =0.64m

then, replacing values, we get:

(425N)(0.64m) = \frac{1}{2}(43.36kg)v^2

Solving for v:

V = 3.54 m/s

7 0
3 years ago
At what depth of lake water is the pressure equal to 201kpa?
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Answer:

20m

Explanation:

Pressure = pgh

p = density of water 1000

kg/m^3

g = acceleration due to gravity 9.81 m/s^2

h is the depth of water

Pressure = 201 kPa = 201 x 10^3 Pa

201 x 10^3 = 1000 x 9.81 x h

201 x 10^3 = 9810h

h = 20.49 m

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3 years ago
A Ping-Pong ball with mass 2.5 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that
dimaraw [331]

Answer:

0.022m or 2.2cm

Expxlanation:

Step 1:

Data obtained from the question. This includes:

Mass (m) = 2.5g = 2.5/1000 = 2.5x10^-3Kg

Tension (T) = 0.029 N

Density (ρ) = 1000 kg/m3

Acceleration due to gravity (g) = 9.81 m/s2

Diameter (d) =?

Step 2:

Finding an expression to calculate the diameter of the ball. This is illustrated below:

Tension = weight displaced - weight of the ball

Weight displaced = Mass of water x acceleration due to gravity

Mass of water = Density x volume

Mass of water = ρxV

Weight displaced = ρxVxg = ρVg

Weight of the ball = Mass of the ball x acceleration due to gravity

Weight of the ball = mg

Therefore,

Tension = weight displaced - weight of the ball

T = ρVg - mg

Make V the subject of the formula

T = ρVg - mg

T + mg = ρVg

Divide both side by ρg

V = ( T + mg) /ρg. (1)

Recall that the ball is spherical in shape and the Volume of a sphere is given by

V = 4/3πr^3

Radius (r) = diameter (d) /2

V = 4/3π(d/2)^3

V = 4/3πd^3/8

V = πd^3 /6

Substituting the value of V into equation 1, we have

V = ( T + mg) /ρg

πd^3 /6 = ( T + mg) /ρg.

Making d the subject of the formula, we have:

πd^3 /6 = (T + mg) /ρg.

d^3 = 6(T + mg) /πρg.

Taking the cube root of both sides

d = [6(T + mg) /πρg]^1/3

Step 3:

Determination of the diameter of the ball. This is illustrated below:

T = 0.029 N

m = 2.5x10^-3Kg

g = 9.81 m/s2

ρ = 1000 kg/m3

d =?

d = [6(T + mg) /πρg]^1/3

d = [6(0.029 + 2.5x10^-3x9.81)/ πx1000x9.81]^1/3

d = 0.022m

Therefore, the diameter of the ball is 0.022m or 2.2cm

6 0
3 years ago
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