A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 40.4 Hz, determine the first three overtones. Use 343 m/s as the speed of sound in air.
If the speed of sound is 337 m/s, determine the length of an open tube (open at both ends) that has a fundamental frequency of 233 Hz and a first overtone frequency of 466 Hz.

Answer:

hypothesis , hope it helps

Explanation:

Answer:

Inference

Explanation:

Inference is something you predict after testing that's a result after an hypothesis has been made. Hypothesis is an intelligent guess based on some observed phenomena which can be subjected to further testing.

Answer: [tex]2m/s^2[/tex]

Explanation:

[tex]Formula: F=ma[/tex]

Where;

F = force

m = mass

a = acceleration

Solve for a;

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{30N}{15kg}\\ a=2m/s^2[/tex]

Answer:

F = ILB

Explanation:

To find the net force on the conducting bar you take into account the following expression:

[tex]\vec{F}=I( \vec{L}X \vec{B})[/tex]

I: current in the conducting bar

L: length of the bar

B: magnitude of the magnetic field

In this case the direction of the magnetic field and the motion of the bar are perpendicular between them. The direction of the bar is + i, and the magnetic field poits upward + k. The cross product of these vector give us the direction of the net force:

+i X +k = +j

The direction of the force is to the right and its magnitude is F = ILB

Answer:

(a) [tex]emf_L=-LI_{max}\omega cos(\omega t)[/tex]

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

[tex]I(t)=I_{max}sin(\omega t)[/tex]   (1)

(a) The induced emf is given by the following formula

[tex]emf_L=-L\frac{dI}{dt}[/tex]    (2)

You derivative the expression (1) in the expression (2):

[tex]emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)[/tex]

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

[tex]emf_L=-\omega LI_{max}[/tex]

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

At t zero,  the current through the inductor neither increasing nor decreasing because current is zero.

The current through an inductor of inductance L can be calculated by

[tex]\bold {I_t = I_m_a_x sin (\omega t)}[/tex].........1  

(a) The induced emf can be calculated by  

[tex]\bold {emf_L = - L \dfrac {dI}{dt}}[/tex]............2  

Derivative the equation (1) in the equation (2)

[tex]\bold {emf _L= -L \dfrac {d (I _m_a_x sin (\omega t)} {dt}}\\\\\bold {emf _L= -L (I _m_a_x \omega cos( \omega t) }[/tex]

(b) At t=0 the current is zero,

(c) At t = 0 the emf is:

[tex]\bold {emf_L = -\omega LI _m_a_x}[/tex]  

Therefore, at t zero,  the current through the inductor neither increasing nor decreasing because current is zero.

To know more about inductance,

https://brainly.com/question/17431388

Answer:

Explanation:

The power rating of the battery isn't provided. But let us assume that it is one of the common batteries with ratings of 12 V and 50 A.h

Potential energy possessed by water at that height = mgh

m = mass of the water = ρV

ρ = density of water = 1000 kg/m³

V = volume of water = ?

g = acceleration due to gravity = 9.8 m/s²

h = height of water = 50 cm = 0.5 m

Potential energy = ρVgh = 1000 × V × 9.8 × 0.5 = (4900V) J

Energy of the battery = qV

q = 50 A.h = 50 × 3600 = 180,000 C

V = 12 V

qV = 180,000 × 12 = 2,160,000 J

Energy = 2,160,000 J

At a 100% conversion rate, the energy of the water totally powers the battery

(4900V) = (2,160,000)

4900V = 2,160,000

V = (2,160,000/4900)

V = 440.82 m³

Hence, with our assumed power ratings for the battery (12 V and 50 A.h), 440.82 m³ of water at the given height of 50 cm would power the battery.

Incase the power ratings of the battery in the complete question is different, this solution provides you with how to obtain the correct answer, given any battery power rating.

Hope this Helps!!!

Answer:

27J

Explanation:

From conservation of Thermal energy, the total internal energy is the total sum of energy supplied or taken from the system plus work done for or on the system.

Now the change in internal energy would be the sum of the received energy substended in the gas plus the work done by the system which is workdone that it will sustend in pushing the lid. This is expressed mathematically as;

U = Q + (F×d);

U- change in internal energy

Q is the energy received by the system and is positive when energy is received by the system.

Fxd is the workdone and is positive since the gas pushes up the lid- the system does work.

U=12+(3×5)= 27J

Answer:2000

Explanation:

Answer:

it represents a fundamental difference. (more info below)

Explanation:

Production is incessantly developing and expanding in socialist countries, and employment is guaranteed for the entire productive population. Consequently, the relative overpopulation problem has been eliminated. This represents the fundamental difference between socialism's demographic law and capitalism's law.

hope this helped!

Answer:

A. F = 0.06 N

B. t = 6.37 s

Explanation:

A)

First we need to find the constant acceleration of the cart. For this purpose, we use 2nd equation of motion:

s = (Vi)(t) + (0.5)at²

where,

s = distance traveled = 1.62 m

Vi = 0 m/s   (Since, it starts from rest)

t = Time Taken = 4.34 s

a = acceleration = ?

Therefore,

1.62 m = (0 m/s)(4.34 s) + (0.5)(a)(4.34 s)²

1.62 m/9.4178 s² = a

a = 0.172 m/s²

Now, from Newton's Second law, we know that:

F = ma

where,

F = Net Force of the combination = ?

m = Mass pf combination = 354 g = 0.354 kg

Therefore,

F = (0.354 kg)(0.172 m/s²)

F = 0.06 N

B)

Now, for the same force, but changed mass = 762 g = 0.762 kg, we have the acceleration to be:

F = ma

a = F/m

a = 0.06 N/0.762 kg

a = 0.08 m/s²

Now, using  2nd equation of motion:

s = (Vi)(t) + (0.5)at²

1.62 m = (0 m/s)(t) + (0.5)(0.08 m/s²)t²

t² = 1.62 m/(0.04 m/s²)

t = √40.54 s²

t = 6.37 s

Answer:

Approximately [tex]0.608[/tex] (assuming that [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex].)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

This answer will approach this question in two steps:

For simplicity, let [tex]a_{T}[/tex] represent the tangential acceleration ([tex]1.90\; \rm m \cdot s^{-2}[/tex]) of this car.

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to [tex]90^\circ[/tex] or [tex]\displaystyle \frac{\pi}{2}[/tex] radians.

The angular acceleration of this car can be found as [tex]\displaystyle \alpha = \frac{a_{T}}{r}[/tex]. ([tex]a_T[/tex] is the tangential acceleration of the car, and [tex]r[/tex] is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ([tex]u[/tex] and [tex]v[/tex]) to (tangential) acceleration [tex]a_{T}[/tex] and displacement [tex]x[/tex]:

[tex]v^2 - u^2 = 2\, a_{T}\cdot x[/tex].

The idea is to solve for the final angular velocity using the angular analogy of that equation:

[tex]\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta[/tex].

In this equation, [tex]\theta[/tex] represents angular displacement. For this motion in particular:

Solve this equation for [tex]\omega(\text{final})[/tex] in terms of [tex]a_T[/tex] and [tex]r[/tex]:

[tex]\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}[/tex].

Let [tex]m[/tex] represent the mass of this car. The centripetal force at this moment would be:

[tex]\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}[/tex].

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, [tex]m\, g[/tex].

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let [tex]\mu_s[/tex] denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

[tex]F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g[/tex].

The size of this force should be equal to that of the centripetal force when the car is about to skid:

[tex]\mu_s\, m\, g = \pi\, m\, a_{T}[/tex].

Solve this equation for [tex]\mu_s[/tex]:

[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g}[/tex].

Indeed, the expression for [tex]\mu_s[/tex] does not include any unknown letter. Let [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex]. Evaluate this expression for [tex]a_T = 1.90\;\rm m \cdot s^{-2}[/tex]:

[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608[/tex].

(Three significant figures.)

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it =  λ dx where  λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x  / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

=  k λ dx  .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫  k λ  .15  / (x² + .15² )³/² dx

=  k λ  x L / .15 √( L² / 4 + .15² )

b) The length of the rod:

[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )[/tex]

Given:

d = 1.5 mλ = 2.5 nC/m

Let the plastic rod extends from - L to + L .Consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .It will create a field at point P on y -axis.

Distance of point P =[tex]\sqrt{x^2 + 0.15^2}[/tex]

E.F at P due to small charged length[tex]dE = \frac{ k \lambda x.dx}{(x^2 + .15^2 )}[/tex]

Its component along Y - axis = dE cosθ where θ is angle between direction of field dE and y axis

[tex]= \frac{dE x .15 }{\sqrt{x^2 + .15^2} }\\\\= \frac{k \lambda dx .15}{(x^2 + .15^2 )^{1/2}}[/tex]

If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . We can say that the component of field in perpendicular to y axis will cancel out each other.

Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )}[/tex]

Find more information about Electric field here:

brainly.com/question/14372859

Soundproofing material is required for blocking sound during some works like recording voice in the studio. Image D represents the interaction of a sound wave with soundproofing material in a recording studio.

Soundproofing is done by absorbing the sound. A very much used material for this is a dense foam.

Foam and like materials absorbs sound and it travels directly into the soft surface resulting in soundproofing.

Thus, the correct option is C, as the D image is showing the absorption.

For more details regarding soundproofing, visit:

https://brainly.com/question/8980142

#SPJ2

Answer: C.D

Explanation:...

Answer:

32mi

Explanation:

If 1lb contains 3,500 Cal

It means the number of hours required to burn 3500cal would be;

3500/331 = 10.57hours

But a brisk walk is 3.0 mi/h,

It means a distance of 3.0 × 10.57 mi would be covered = 31.71 miles

32miles{ approximated to the nearest whole}

Note Distance = speed × time

Answer:

The distance covered by puck A before collision is  [tex]z = 8.56 \ m[/tex]

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  [tex]v_A = 3.90 \ m/s[/tex]

        The speed of puck B is  [tex]v_B = 4.30 \ m/s[/tex]

The distance covered by puck A is mathematically represented as

     [tex]z = v_A * t[/tex]

  =>  [tex]t = \frac{z}{v_A}[/tex]

 The distance covered by puck B  is  mathematically represented as

      [tex]18 - z = v_B * t[/tex]

=>   [tex]t = \frac{18 - z}{v_B}[/tex]

Since the time take before collision is the same

        [tex]\frac{18 - z}{V_B} = \frac{z}{v_A}[/tex]

substituting values

          [tex]\frac{18 -z }{4.3} = \frac{z}{3.90}[/tex]

=>      [tex]70.2 - 3.90 z = 4.3 z[/tex]

=>       [tex]z = 8.56 \ m[/tex]

Answer:

J = 0.800 kg m/s

Fmax = 291 N

Explanation:

During the fall, energy is conserved.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × 9.81 m/s² × 1.45 m)

v = 5.33 m/s

Alternatively, you can use kinematics to find the velocity.

Impulse = change in momentum

J = Δp

J = mΔv

J = (0.150 kg) (5.33 m/s)

J = 0.800 kg m/s

Impulse = area under F vs t graph

J = ∫ F dt

J = ½ Fmax Δt

(0.800 kg m/s) = ½ Fmax (0.0055 ms)

Fmax = 291 N

Answer:

by using it's buoyant or floating effect by Archimedes.

the buoyant force act on the astronauts body and make he/ she feels like in low gravity.

the buoyant force equation is

F = Density of liquid x earth gravitational field x volume of astronauts body and suit.

the Weight of astronauts in the pools will be less than in the land or air.

Weight in water = weight in air/land - buoyant force

so the astronauts will feel like in the outer space with low gravity.

Answer:

Letter A. [tex]y=4.55 m[/tex]

Explanation:

Let's use the wave equation:

[tex]y=Asin(kx-\omega t)[/tex]

Therefore the displacement y will be:

[tex]y=6.44*sin(2.34*1.21-2.88*0.71)[/tex]

[tex]y=4.55 m[/tex]

The answer is letter A.

I hope it helps you!

Answer:

Explanation:

Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.

Amplitude (A) of the simple harmonic wave = 6.44 m

wave number (k) of the given wave = 2.34 m-1

Angular frequency (ω) of the given wave = 2.88 rad/s

Displacement x = 1.21 m and time t = 0.71 s

Then the general equation for the displacement of the given simple harmonic wave at given x and time t is given by

y = Asin(kx - ωt)

= (6.44 m)sin[(2.34 m-1)(1.21 m) - (2.88 rad/s)(0.71 s)]

Y=6.44sin(0.7866 rad)

0.7866rad*(180 degrees/pi rad) =45.1

Y=6.44sin(45.1)

Y=4.55m

Answer:

D

Explanation:

Insoluble impurities would not change the constituent of the substance. Soluble would for example salt water takes longer time for the water to become vapour when subjected to the same temperature that normal water.

Wind would affect, the more windy the tendency for particles of the liquid to be moved into the atmosphere.

With an increase in surface area, the evaporation rate increase . Take a clue from water placed on the ground and exposed to the atmosphere and that same quantity of water is placed in a cup. That on the floor would evaporate faster.

Similarly the higher the temperature a substance is subjected to the easier is it's rate of evaporation. Take for instance water in a cup placed in the sun and that same placed in a room with mild temperatures than that of the sun.With time that in the sun decreases in volume faster than that in the room.

Answer:

786 Hz

Explanation:

Recall, the speed of sound is

v = 332 + 0.6t

Where t = 23°

v = 332 + 0.6(23)

v = 332 + 13.8

v = 345.8 m

Also, it is known that distance between two consecutive resonance length is half of the wavelength.

L2 - L1 = λ/2

34 - 12 = λ/2

λ/2 = 22

λ = 44 cm

Finally, remember that also

Frequency = speed/ wavelength

Frequency = 345.8/0.4

Frequency = 786 Hz

Therefore, the frequency of the tuning fork is 786 Hz

Answer:

a. 63.89°  in the north-southward manner

b.  2.2 sec

Explanation:

The goose is flying at 100 km/h

Air from east to west is 44 km/h

angle relative to the north-south direction for the bird to travel south will be

cos∅ = 44/100 = 0.44

∅ = [tex]cos^{-1}[/tex]0.44 = 63.89°  in the north-southward manner

Speed south relative to the ground will be v

Tan 63.89 = v/100

2.04 = v/100

v = 2.04 x 100 = 204 km/hr

to cover a distance of 450 m from north to south at this speed time will be

t = d/v = 450/204 = 2.2 sec

Answer:

C

Explanation:

The intermolecular forces between the water molecule is less binding than that of the copper molecule. Hence the water would take a shorter time to be converted to vapour where the temperature of boiling is constant however the temperature of that of the copper molecule keeps increasing.

Answer:

Everytime you do an experiment you need something that is regular. For example if you try and measure how much germs spread in bread. you need 1 bread thats clean and 3 different breads for different molds. So thats called a CONTROL

AAAAAAAAAAAA is the answer

Answer:

55.3

Explanation:

The computation of the number of bright-dark-bright fringe shifts observed is shown below:

[tex]\triangle m = \frac{2d}{\lambda} (n - 1)[/tex]

where

d = [tex]3.95 \times 10^{-2}m[/tex]

[tex]\lambda = 400 \times 10^{-9}m[/tex]

n = 1.00028

Now placing these values to the above formula

So, the  number of bright-dark-bright fringe shifts observed is

[tex]= \frac{2 \times3.95 \times 10^{-2}m}{400 \times 10^{-9}m} (1.00028 - 1)[/tex]

= 55.3

We simply applied the above formula so that the number of bright dark bright fringe shifts could come

Answer:

Some of the advantages if our body had magnetic properties are as follows:

Hence, magnetic properties are somehow beneficial for humans.  

Answer:

Overall charge still remains zero on conductor until touched by charged rod.

Explanation:

Here, we want to know what has happened to the overall charge on the conductor.

Since the conductor is neutral, the overall charge on the conductor must remain zero because positive charge is induced on close end to rod then equal and negaitve charge is induced on far end to rod.

Thus, overall charge still remains zero on conductor until touched by charged rod.

Answer:

a = 17 m / s²

Explanation:

For this experiment that the astronauts are carrying out, the value of the relation of gravity is cosecant, therefore we can use the kinematic relations

         v² = v₀² + 2a y

They indicate the initial speed 1.2 m / s the final speed 4.3 m / s and the distance remembered 0.5 m

we clear

        a = (v² - v₀²) / 2y

we calculate

       a = (4.3² -1.2²) / 2 0.5

       a = 17 m / s²

this is the gravity of the new planet

Answer:

B. At the instant when the block is at the highest point, directed at the block.

Explanation:

Motion of an object is the change in the position of the object with respect to time. On the earth, gravity has a great influence on the motion of an object (especially in a vertical direction).

When the block is projected up in the air, it moves with a varying velocity until the velocity becomes zero due to gravity. Which make the object to rest a little in the air (when velocity = gravity) and starts to fall freely.

To ensure hitting the block by the ball, it is thrown at the block when the block is at its highest point in the air. Since the block would be at rest at this instant before it start to fall at a constant acceleration under gravity.

Answer:

U = 218 nJ

Explanation:

We are given;

Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m

Voltage across the capacitor; V = 96 V

Dimension of the square plates is 7.2cm x 7.2cm.

So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²

Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²

From relative permeability table;

Dielectric constant of Pyrex; k1 = 5.6

Dielectric constant of polystyrene; k2 = 2.56

Now, formula for capacitance of a capacitor with Dielectric is;

C = kC_o

Where, C_o = ε_o(A/d)

Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)

Since we have 2 capacitor, thus ;

C1 = k1*ε_o*(A/d)

C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C1 = 1.51 × 10^(-10) F

Similarly;

C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C2 = 0.691 × 10^(-10) F

For capacitors in series, formula for total capacitance(Cs) is;

1/Cs = (1/C1) + (1/C2)

Simplifying this, we have;

Cs = (C1*C2)/(C1 + C2)

Plugging in the relevant values ;

Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))

Cs = 0.474 × 10^(-10) F

The formula for energy stored in a capacitor with 2 Dielectrics is given as;

U = ½Cs*V²

So,

U = ½ × 0.474 × 10^(-10) × 96²

U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ

Answer:

a.[tex]5.66ms^{-2}[/tex]

b.55 m

Explanation:

We are given that

Mass ,m=0.9 kg

Length of string,l=50 cm=[tex]\frac{50}{100}=0.50 m[/tex]

1 m=100 cm

[tex]\theta=30^{\circ}[/tex]

R=55 m

a.Centripetal acceleration

[tex]a_c=gtan\theta[/tex]

[tex]a_c=9.8tan30^{\circ}[/tex]

[tex]a_c=5.66 m/s^2[/tex]

Hence, the radial acceleration of the bus=[tex]5.66m/s^2[/tex]

b. Radius of curve,R=55 m

Answer:

47.8rad/s

Explanation:

For energy to be conserved.

The potential energy sustain by the object would be equal to K.E

P.E = m× g× h = 2 × 9.81× 3.5= 68.67J

Now K.E = 1/2 × I × (w1^2 - w0^2)

I = 2/3 × M × R2

= 2/3 × 2 × (0.23)^2= 0.0705

Hence

W1 = final angular velocity

Wo = initial angular velocity

From P.E = K.E we have;

68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)

(w1^2 - w0^2) = 1948.09

W1^2 = 1948.09 + (18.3^2)

W1^2=2282.98

W1 = √2282.98

=47.78rad/s

= 47.8rad/s to 1 decimal place.