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Reil [10]
2 years ago
11

The fundamental frequency of a resonating pipe is 150 Hz, and the next higher resonant frequencies are 300 Hz and 450 Hz. From t

his information, what can you conclude
Physics
1 answer:
olasank [31]2 years ago
3 0

Answer:

Explanation:

In case of open organ pipe ,

for   fundamental note   2l = wavelength

frequency n = V / Wave length , V is velocity of sound.

f₁ = V / 2I

for   first overtone  note   2l /2  = wavelength

wavelength = l

frequency = V / wavelength

f₂ = V / l

for   second  overtone  note   2l /3  = wavelength

wavelength = 2l /3

frequency = V / wavelength

f₃ =  3V /2l

f₁ : f₂ : f₃ : : 1 : 2 : 3

In case of closed organ pipe ,

for fundamental note   l = 4 x wavelength

frequency n = V / Wave length , V is velocity of sound.

f₁ = V / 4 I

for   first overtone  note   l /  =   3 wavelength / 4

wavelength = 4 l / 3

frequency = V / wavelength

f₂ =3 V /4 l

for   second  overtone  note   l   =  5 wavelength /4

wavelength = 5l /4

frequency = V / wavelength

f₃ =  5 V /4 l

f₁ : f₂ : f₃ : : 1 : 3 : 5.

In case of open organ pipe , frequencies are in any integral ratio.

In case of closed organ pipe , frequencies are only in odd multiple

Since the given ratio are in any integral , the organ pipe must be open organ pipe.

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A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at
GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0
1 year ago
Because the soles of your shoes have cleats, you can exert a forward force of 100 N even on slippery ice. A 10-kg picnic cooler
Brilliant_brown [7]

Answer:

you must throw 3 snowballs

Explanation:

We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment

Initial. Before bumping that refrigerator

          p₀ = n m v₀

Where n is the snowball number

Final. When the refrigerator moves

         pf = (n m + M) v

The moment is preserved because the forces during the crash are internal

        n m v₀ = (n m + M) v

        n m (v₀ - v) = M v

        n = M/m    v/(vo-v)

Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton

          F = m a

          a = F / m

The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm

          v₁² = v₀² + 2 a x

          v₁² = 0+ 2 (100/1) 1

          v₁ = 14.14 m / s

This is the initial speed for the crash

         v₀ = v = 14.14 m / s

  Let's calculate

           n = M/m   v/ (v₀-v)

           n = 10/1   3 / (14.14 -3)

          n = 2.7 balls

you must throw 3 snowballs

7 0
2 years ago
A plane took 7 hours to travel 2020 km. For the first 4 hours,
kicyunya [14]

Answer:

-6.6 km/h

Explanation:

In 7hr plane travelled 2020km;

For the first 4hr the average speed was 310km/h;

d=st, s=d/t;

Distance covered in first 4h is d = 310km/h×4h = 1240km;

See the image attached for further solution

8 0
2 years ago
Read 2 more answers
A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over
kifflom [539]

Answer:

f_D = =3.24 N/m

Explanation:

data given

properties of air\nu\ of air =19.31*10^{-6} m2/s

\rho = 1.048 kg/m3

k = 0.0288 W/m.K

WE KNOW THAT

Reynold's number is given as

Re =\frac{VD}{\nu}

      = \frac{ 15*0.025}{19.31*10^{-6}}

      = 1.941 *10^4

drag coffecient is given as

C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}

solving for f_D

f_D = C_D A_f*\frac{\rho v^2}{2}

     =C_D D*\frac{\rho v^2}{2}

Drag coffecient for smooth circular cylinder is 1.1

therefore Drag force is

f_D = 1.1*0.025 *\frac{1.048*15^2}{2}

f_D = =3.24 N/m

4 0
2 years ago
An airplane weighing 5000 lb is flying at standard sea level with a velocity of 200 mi/h. At this velocity the L/D ratio is a ma
saul85 [17]

Answer:

98.15 lb

Explanation:

weight of plane (W) = 5,000 lb

velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s

wing area (A) = 200 ft^{2}

aspect ratio (AR) = 8.5

Oswald efficiency factor (E) = 0.93

density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}

Drag = 0.5 x ρ x v^{2} x A x Cd

we need to get the drag coefficient (Cd) before we can solve for the drag

Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)

where

  • induced drag coefficient (Cdi) = \frac{Cl^{2} }{n.E.AR} (take note that π is shown as n and ρ is shown as p)    

        where lift coefficient (Cl)= \frac{2W}{pAv^{2} }=\frac{2x5000}{0.002377x200x293.3^{2} } = 0.245

        therefore

       induced drag coefficient (Cdi) = \frac{Cl^{2} }{n.E.AR} = \frac{0.245^{2} }{3.14x0.93x8.5} = 0.0024

  • since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)
  • Cd = 0.0024 + 0.0024 = 0.0048

Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.        

 Drag = 0.5 x ρ x v^{2} x A x Cd

Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb

8 0
2 years ago
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