The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.
1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?
4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third
harmonic.

Answer:

2)The Cell with OER 7

Explanation:

OER is the acronym for Oxygen Enhancement Ratio. It is the measure of the  enhancement of the effect of ionizing radiation due to the presence of oxygen. The ionization effect can be detrimental or therapeutic (use in cancer treatment). OER is the ratio of radiation dose during the lack of oxygen (hypoxia), to the dosage in the presence of oxygen (air is used as a reference). From the definition, one can see that the higher the OER the higher the sensitivity of the cell.

A) The acceleration of the electron along the x -axis is ; 4.57 * 10⁻¹¹ m /s²

B) The speed that would result in the electron experiencing an acceleration  along the x-axis is 4.57 * 10⁻¹¹  * time  m/s

Given Data :

Electric field ( E ) = ( 2.60i + 5.90j ) V/m

Magnetic field ( B ) = 0.400 k T

Velocity ( v ) = 8.0i m/s

Applying Lorentz force

F = q ( E + ( v * B ) )

  = 1.6 * 10⁻¹⁹ ( 2.60 i  +  5.90 j  + ( 8.0 i * 0.4 k ) ) N

  = 1.6 * 10⁻¹⁹ ( 2.60 i  +  5.90 j + ( 3.2 ( -j ) ) N

  = 1.6 * 10⁻¹⁹ (  2.60 i  + 2.70 j ) N

Ax = 4.57 * 10⁻¹¹ m /s²

Ax = Fx / Mc

    = ( 1.6 * 10⁻¹⁹ * 2.60 ) / 9.1 * 10⁻³¹

    = ( 4.16 * 10⁻¹⁹ ) / 9.1 * 10⁻³¹

    = 0.457 * 10¹²

    = 4.57 * 10⁻¹¹ m /s²

Therefore The speed that would result in the electron experiencing an acceleration  along the x-axis is 4.57 * 10⁻¹¹  * time

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Answer:

[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].

The unit of both sides of this equation are [tex]\rm s[/tex].

Explanation:

The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].

The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].

On the right-hand side of this equation:

[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].

Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].

Answer:

the net work is zero

Explanation:

Work is defined by the expression

        W = F. ds

Bold type indicates vectors

In this problem, the friction force does not decrease, therefore it will be zero.

Consequently for work on a closed path it is zero.

The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero

Answer:

A.

Explanation:

Amplitude measures how much a wave rises or falls. This is illustrated by A.

In the diagram, the amplitude of the wave is shown by A.

There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.

The amplitude of a variable is simply a measure of change relative to its central position. In contrast, magnitude is a measure of the distance or quantity of a variable irrespective of its direction.

Amplitude is a property that is unique to waves and oscillations.

Therefore, in the diagram, the amplitude of a wave is shown by A.

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Answer:

The net energy transferred between two objects

Explanation:

The physical property of matter that expresses hot or cold is called temperature. It demonstrates the thermal energy. A thermometer is used to measure temperature. It defines the rate to which the chemical reaction occurs. It tells about the thermal radiation emitted from an object.

The correct option that defines temperature is option C.

Answer:

A measure of the movement of atoms or molecules within an object

Explanation:

Process of elimination

Answer:

[tex]\boxed{\sf 65.3 \ inches}[/tex]

Explanation:

1 foot = 12 inches

Sammy is 5 feet tall.

5 feet = ? inches

Multiply the feet value by 12 to find in inches.

5 × 12

= 60

Add 5.3 inches to 60 inches.

60 + 5.3

= 65.3

Answer:

3.9E-8

Explanation:

We know that

Mv²/r = Bqv

So

r= mv/Bq

But E is 1/2mv² which is 5.53eV

m²v² =2m x 5.53eV

mv = √( 2 x 9.1E-31)

So

r= √( 2 x9.1E-31 x 5.53)/ 7.83x10^-4 x1.6E-19

= 3.9x10-8cm

​

Explanation:

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

Answer:

the interation blood veins

Explanation:

Answer:

Explanation:

a )

focal length of convex spherical mirror

f = 36/2  cm = 18 cm

object distance u = - 25 cm

mirror formula

[tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]

[tex]\frac{1}{v} + \frac{1}{- 25} = \frac{1}{18}[/tex]

[tex]\frac{1}{v} = \frac{1}{25} + \frac{1}{18}[/tex]

v = 6.28  cm .

It is positive hence the image will be erect / upright  and formed on the back of the mirror.

For object distance of 47 cm

u = - 47 cm

Putting the values in the mirror formula

[tex]\frac{1}{v} + \frac{1}{- 47} = \frac{1}{18}[/tex]

[tex]\frac{1}{v} = \frac{1}{ 47} + \frac{1}{18}[/tex]

v = 13  cm

It is positive hence the image will be erect / upright and formed on the back of the mirror.

The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %

The known values are;

The increase in pressure per 10 meter increase in depth = 1.0 atm

The depth of the deepest ocean = 12 km = 12,000 m

The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm

The unknown

The percentage the density of water increased in the deepest ocean

Strategy;

Find the pressure at the deepest point of the deepest ocean and apply the compressibility

We have;

[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]

The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm

Therefore, we have for one cubic meter of water

[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]

Therefore;

[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³

The new volume = V - [tex]\mathbf{\partial}[/tex]V

∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³

The initial density = mass/(1 m³)

The new density = mass/(0.96 m³)

The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;

[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]

∴  [tex]\mathbf{\partial}[/tex]ρ% =  4.1[tex]\mathbf {\overline 6}[/tex] %

The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %

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Answer:

The uncertainty in momentum is 5.25x 10^25Jsm

Explanation:

We know that

h bar = h/2π

So

1.05x 10^34=h/2pπ

h=1.05x 10^ 34(2π)=6.597x 10^-34Js

dp=(6.597x10^-34/4pπ)/(1x10^-10)

=5.25x10^-25 Jsm

Answer:

Sweat also contains ammonia and urea, which are produced by the body when it breaks down proteins from the foods you eat.

Hope this helps..

Answer:

Scientific models are representations of objects, systems or events and are used as tools for understanding the natural world. Models use familiar objects to represent unfamiliar things. Models can help scientists communicate their ideas, understand processes, and make predictions.

Answer:

hi there

Explanation:

1 - III

2- 1

3-1

HOPE IT HELP YOU

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Explanation:

1 . 3

2. 1

3. 2

I hope it is helpful to you.

Answer:

The near point is  [tex]n =44.8 \ cm[/tex]

Explanation:

From the question we are told that

   The power is  [tex]P = 1.50[/tex]

   The  distance from the eye is  [tex]k = 1.8 \ cm[/tex]

    The distance of the book from the eye is [tex]z = -28 \ cm[/tex]

Generally the focal length of the glasses is  

       [tex]f = \frac{1}{P}[/tex]

=>   [tex]f = \frac{1}{1.50 }[/tex]

=>   [tex]f = 0.667 \ m[/tex]

=>   [tex]f = 66.7 \ cm[/tex]

The object distance is evaluated as

     [tex]u = z + k[/tex]

=>   [tex]u = -28 + 1.8[/tex]

=>  [tex]u = -26.2 \ cm[/tex]

The image distance is evaluated from lens formula as

       [tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]

=>   [tex]\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}[/tex]

=>   [tex]v=- \frac{1}{0.0232}[/tex]

=>    [tex]v=- 43 \ cm[/tex]

The  near point is evaluated as

      [tex]n = -v + k[/tex]

=>    [tex]n =-(-43) + 1.8[/tex]

=>    [tex]n =44.8 \ cm[/tex]

Answer:

Hey there!

This can be a confusing topic, so it's totally fine if you get confused...

First, Seismic Attenuation is how seismic waves lose energy as they expand and spread.

Secondly, when distance increases, amplitude decreases. This is because the distance (spherical spreading would mean radius) is inversely proportional to amplitude.

Let me know if this helps :)

Answer:

i) 545.2 N  upwards

ii) 828.2 N  downwards

Explanation:

mass of the roller = 70 kg

force exerted = 200 N

angle the force makes with the ground ∅ = 45°

weight of the roller W = mg

where

m is the mass of the roller

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

weight of the roller = 70 x 9.81 = 686.7 N

The effective vertical force exerted by the man = F sin ∅ = 200 x sin 45°

==> F = 200 x 0.707 = 141.5 N

i) if the man pulls, then the exerted force will be in opposite direction to the weight of the roller vertically upwards

Resultant vertical force = 686.7 N - 141.5 N = 545.2 N  upwards

ii) if he pushes, then the exerted force will be in the direction of the weight vertically downwards

Resultant vertical force = 686.7 N + 141.5 N = 828.2 N  downwards

Answer:

The initial kinetic energy of the bullet is closest to 491.87 J

Explanation:

Given;

mass of bullet, m₁ = 18g = 0.018kg

mass of block, m₂ = 10kg

height moved by the block, h = 9 mm = 0.009 m

time taken for the bullet to travel through the block, t = 0.001s

let the initial velocity of the bullet = v₁

let the final velocity of the bullet = v₂

Apply the principle of conservation of linear momentum;

initial momentum = final momentum

0.018v₁ = v₂(0.018 + 10)

0.018v₁ = 10.018v₂ -----equation (1)

Apply the law of conservation of energy when the bullet lifts the block through 9mm

mgh = ¹/₂mv₂²

gh = ¹/₂v₂²

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.009)

v₂ = 0.42 m/s

Substitute in v₂ in equation 1, to determine the initial velocity of the bullet;

0.018v₁ = 10.018v₂

0.018v₁  =  10.018(0.42)

0.018v₁  = 4.208

v₁ = 4.208 / 0.018

v₁ = 233.78 m/s

Now, determine the initial kinetic energy of the bullet;

K.E₁ = ¹/₂m₁v₁²

K.E₁ = ¹/₂(0.018)(233.78)²

K.E₁ = 491.87 J

Therefore, the initial kinetic energy of the bullet is closest to 491.87 J

Answer:

C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays

Explanation:

radio waves have lowest  energy , lowest  frequency and highest  wavelength

gamma rays  have highest  energy , highest  frequency and least  wavelength

Answer: C

Explanation:

Answer:

0.99Hz

Explanation:

Using F= -mx ( spring force)

At equilibrium the gravitational force will be balanced by the spring force so mg= kx

K= mg/ 0.25 N/m

But

Frequency f= 1/2pi √g/0.25

Frequency is 0.99Hz

The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz

What information do we have?

Length starched = 2.5 cm

F = Kx

We know that

F = mg

So,

mg = Kx

K/m = g/x

[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }[/tex]

Frequency of oscillation = 3.15 Hz

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Answer:

hhhbbbbbbbbbbbbbbnnnnnbbhb

Answer:

4.5x 10^ -9m

Explanation:

See attached file

Answer:

The radius is  [tex]r_{min} = 0.00226 \ m[/tex]

Explanation:

   From the question we are told that

      The  elastic limit(stress) is [tex]\sigma = 5.0*10^{8} \ N /m^2[/tex]

      The length is  [tex]L = 4.0 \ m[/tex]

      The weight of the commercial sign is    [tex]F_s = 8000 \ N[/tex]

       The maximum extension of the wire is  [tex]\Delta L = 5.0 \ cm = 0.05 \ m[/tex]

Generally the elastic limit of an alloy (stress) is is mathematically represented as

            [tex]\sigma = \frac{ F_s }{ A }[/tex]

Where A is the cross-sectional area of the wire which is mathematically represented as

         [tex]A = \pi r^2[/tex]

here [tex]r = r_{min}[/tex] which is the minimum radius of the wire that support the commercial sign

So

          [tex]\sigma = \frac{ F_s }{ \pi r_{min}^2 }[/tex]

=>       [tex]r_{min} = \sqrt{\frac{F_s}{\sigma * \pi} }[/tex]

substituting values

             [tex]r_{min} = \sqrt{\frac{8000}{ 5.0* 10^8 * 3.142} }[/tex]

           [tex]r_{min} = 0.00226 \ m[/tex]

As No diagram attached I am taking all are connected in series

We know

[tex]\boxed{\sf C_{eq}=C_1+C_2\dots}[/tex]

[tex]\\ \sf \longmapsto C_{eq}=12+12+7+15+15+15[/tex]

[tex]\\ \sf \longmapsto C_{eq}=24+7+45[/tex]

[tex]\\ \sf \longmapsto C_{eq}=76\mu F[/tex]

Answer:

Compare the volume  (weight) of CO2  gas produced to the weights of sugar and yeast that were used to produce the CO2. Ideally, both the yeast and sugar would be entirely consumed.

Answer:

b. Third Law entropy  

Explanation:

Third law entropy: In physics, the term "third law entropy" or "the third law of thermodynamics" states that the specific entropy of a particular system at "absolute zero" is considered as a "well-defined constant". It occurs because any system at "zero temperature"  tends to exists or persists in its "ground state" in order for the entropy to be determined or described only by the "degeneracy" of the given ground state.

In the question above, the correct answer is option b.

CHECK COMPLETE QUESTION BELOW

you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?

A) four times more than you do on Earth.

B) two times less than you do on Earth.

C) the same as you do on Earth

D) two times more than you do on Earth

Answer:

OPTION C is correct

The same as you do on Earth

Explanation :

According to law of gravitation :

F=GMm/R^2......(a)

F= mg.....(b)

M= mass of earth

m = mass of the person

R = radius of the earth

From law of motion

Put equation b into equation a

mg=GMm/R^2

g=GMm/R^2

g=GM/R^2

We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have

m= 4M

r=(2R)^2=4R^2

g= G4M/4R^2

Then, 4in the denominator will cancel out the numerator we have

g= GM/R^2

Therefore, g remain the same

Answer:

M = F/3μ g - M₁/3

Explanation:

To solve this exercise we must use the equilibrium conditions translations

         ∑ F = 0

In the attachment we can see a free body diagram of each block

Block M (upper)

X axis

      fr₁ + F₂ -F = 0

      F = fr₁ + F₂              (1)

axis

     N₁-W = 0

     N₁ = Mg

the friction force has the formula

     fr₁ = μ N₁

     F = μ Mg + F₂

bottom block

X axis

     F₂ - fr₁ - fr₂ = 0

     F₂ = fr₁ + fr₂

Y axis

     N - W₁ -W = 0

     N = g (M + M₁)

we substitute

       F₂ = μ Mg + μ (M + M1) g

       F₂ = μ g (2M + M₁)

we substitute in 1

      F = μ M g + μ g (2M + M₁)

      F = μ g (3M + M₁)

we look for mass M    

      M = (F -  μ g M₁)/ 3μ g

      M = F/3μ g - M₁/3

the exercise does not have numerical data