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How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenl

spin [16.1K]

Answer:

L=0.654 m

Explanation:

Concepts and Principles

1- The speed of sound in air is expressed as a function of the temperature of air as follows:

v=(331 m/s)√(1+T_C/273°C) (1)

where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.

Standing Wave Patterns in Pipes:

A pipe open at both ends can have standing wave patterns with resonant frequencies:

f=v/λ=nv/2L n=1,2,3.........

where v is the speed of sound in air.

Given Data

f_1 (fundamental frequency of the flute) = 262 Hz

T (temperature of the air) = 20°C

The flute is open at both ends.

Required Data

We are asked to determine the length of the tube.

Solution

The speed of sound in air at temperature T = 20°C is found from Equation (1):

v=(331 m/s)√(1+T_C/273°C)

=342.91 m/s

The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):

f=v/2L

Solve for L:

L=v/2f_1

L=0.654 m

7 0

3 years ago

An acorn falls from a branch located 9.8 m above the ground. After 1 s of falling, the acorn's velocity will be 9.8 m/s downward

goldenfox [79]

Answer:

The acorn hasn't hit the ground because it only falsl half of the branch distance from the ground

Explanation:

given information:

h =9.8

t =1 s

g = 9.8

the average speed

v = 1/2 gt²

= 1/2 (9.8) (1)²

= 4.8 m/s

the distance in 1s

h = v t

= 4.8 (1)

= 4.8 m

the acorn hasn't hit the ground because it only falsl half of the branch distance from the ground

=

8 0

3 years ago

On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The

Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

$a = g (\sin \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2$

Velocity at the bottom

$v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s$

after travelling 4m , its velocity becomes 0

$a=\frac{v^2-u^2}{2s}$

$a=\frac{0-u^2}{2s}$

$a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2$

Coefficient of kinetic friction

μ = F/N

$=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31$

Therefore, the Coefficient of kinetic friction is 0.31

8 0

2 years ago

Read 2 more answers

A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Zielflug [23.3K]

Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to start crate moving is $F_{s} =\mu_{s} mg$ .

but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie $F_{v} =\mu_{k} mg$ .

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.

Magnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

4 0

3 years ago

A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two

natta225 [31]

Answer:

-15 m/s

Explanation:

The computation of the velocity of the 4.0 kg fragment is shown below:

For this question, we use the correlation of the momentum along with horizontal x axis

Given that

Weight of stationary shell = 6 kg

Other two fragments each = 1.0 kg

Angle = 60

Speed = 60 m/s

Based on the above information, the velocity = v is

$1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0$

$\frac{60}{2} + \frac{60}{2} - 4\ v = 0$

$v = \frac{60}{4}$

= -15 m/s

4 0

3 years ago