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Amiraneli [1.4K]
3 years ago
15

Some pipe organs create sounds lower than humans can hear. This ""infrasound"" can still create physical sensations. What is the

fundamental frequency of the sound from an open-open pipe that is 32 feet long (a common size for large organs)? What length open-closed tube is necessary to produce this note? Assume a sound speed of 343 m/s.
Physics
1 answer:
Dominik [7]3 years ago
8 0

Answer: 17.59 Hz and 4.87 m

Explanation:

The fundamental frequency of the sound from an open-open pipe is given as

f= \frac{v}{2L}

where v= 343 m/s

L= 32 feet= \frac{32}{3.281} = 9.75 m

So,

f= \frac{343}{2*9.75} = 17.59 Hz

The length of open-closed tube is related to frequency by formula

f= \frac{v}{4L}

or L=\frac{v}{4f}

L= \frac{343}{4*17.59}

L= 4.87 m

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Answer:

The Celcius and kelvin scale are related unit for unit. One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale. The only difference between these two scales is the zero point.

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2 years ago
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Which statement describes a controlled experiment?
olga_2 [115]

Answer:

In a controlled experiment, an independent variable (the cause) is systematically manipulated and the dependent variable (the effect) is measured; any extraneous variables are controlled. The researcher can operationalize (i.e. define) the variables being studied so they can be objectivity measured.

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3 years ago
Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a cur
Dafna1 [17]

Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 is the vacuum permeability

I_1 is the current in the 1st wire

I_2 is the current in the 2nd wire

r is the separation between the wires

In this problem

I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m

Substituting, we find the force per unit length on the two wires:

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out  from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

3 0
3 years ago
On a 100km track , a train travels the first 30km with a speed of 30km/h . How fast the train travel the next 70 km if the avera
nirvana33 [79]

Solution :-

Given :

Distance 1 = 30 km

Distance 2 = 70 km

We know that speed = distance/time

and, Average speed = total distance/total time taken

When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour

Average speed = 9distance 1 + distance 2)/(time 1 + time 2)

AS time 2 or t2 is time taken for the second part of the journey of 70 km

⇒ 40 = 100/(1 + t2)

⇒ 40 + 40t2 = 100

⇒ 40t2 = 100 - 40

⇒ 40t2 = 60

⇒ t2 = 60/40

⇒ t2 = 1.5

So, t2 or time taken to travel the second part of the journey is 1.5 hours.

Speed of the second part of the journey = distance 2/time 2

⇒ 70/1.5

⇒ 46.666 km/hr or 46.7 km/hr.

Hence the answer is = 46.666 km/hr or 46.7 km/hr.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

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3 0
4 years ago
If eight water waves pass an ocean buoy each minute, and successive wave crests are 20 m apart, find the wave speed:____________
Llana [10]

Answer:

The wave speed is calculated below:

Explanation:

Given,

number of waves passed per minute = 8

time period = 1 minute = 60 s

distance between successive wave crests = 20 m

waves passing interval per second = \frac{8}{60} s^{-1}

Now,

wave speed = 20 m × \frac{8}{60} s^{-1}

                     = \frac{8}3} m/s

                     = 2.67 m/s

Hence the wave speed is 2.67 m/s.

4 0
3 years ago
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