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Elina [12.6K]
3 years ago
7

Can someone help with these?

Physics
1 answer:
Slav-nsk [51]3 years ago
4 0

Twelve

For an open ended pipe, the fundamental frequency is dependent on the length of the pipe and the speed of sound.  Slightly less dramatic limiting factors are the temperature and pressure.

In an open pipe column, the fundamental frequency is f1 = V_sound / (2*L) which means that the longer the pipe, the lower the frequency.

Each frequency thereafter is just a multiple of the fundamental frequency.

f2 = 2* f1

f3 = 3 * f1

fn = n*f1

But that is not really what you are being asked. You are asked about the wavelength

Start with the fundamental formula

v = f wavelenght

f = n * v/2L

v = (n*v/2L) * wavelength  The "v"s cancel out

2L / n = wavelength.

Thirteen

Fourteen

If the pipe increases in length, the frequency will go down and the wavelength will go up


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An inverse-time circuit breaker (CB) is used for branch-circuit short-circuit and ground-fault protection for a 30-horsepower, 2
Maru [420]

Answer:

Explanation:

Motor rating is given in horsepower (hp), it will be converted in watt (W).

Standard to install circuit breaker for an electric circuit is usually 20% ~ 25% more than the Rated Current of the circuit

while

Standard to install overload relay for an electric circuit is usually 20% ~ 25% more than the Running Current of the circuit.

So, to find the maximum capacity of the circuit breaker, rated current of the motor will be multiplied by 1.2 ~ 1.25

Step by Step Explanation:

30hp = 22371W (as 1hp = 745.7)

Assuming unity power factor (cosФ=1) and 208V phase to phase voltage:

Rated Power (watt) = √3 . V.I. cosФ

<em>{if 208V is phase to neutral voltage, then use following formula:</em>

<em> Rated Power (watt) = 3 . V.I. cosФ}</em>

\frac{22371}{\sqrt{3} * 208 * 1} = I

Rated Current = <u>62.169A</u>

So, required maximum rating for circuit breaker is:

20% to 25% of the rated current = 62.17*1.2 ~ 62.17*1.25

=74.6A ~ 77.7A

Hence, any breaker between the above mentioned rating will be appropriate.

3 0
3 years ago
How to tell if something is an electrolyte
schepotkina [342]

Answer: If it has ions, it is an electrolyte

Explanation:

Let's start by explaining that electrolytes are compounds that contain charged particles or<u> ions</u>, which can be cations (positive ions) or anions (negative ions).

So, it is this composition that makes an electrolytic material conduct electricity.

In this sense, the way to identify if a material is an electrolyte or not, is knowing whether it is composed of ions or not.

8 0
3 years ago
How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
3 years ago
An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o
jenyasd209 [6]

Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

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Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

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\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}

\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}

v=-\frac{20}{3}=-6.7 cm

Image distance of apple=-6.7 cm

Magnification=m=\frac{v}{u}=\frac{-\frac{20}{3}}{-20}

Magnification of apple=\frac{1}{3}=0.33

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Which of the following is not an example of centripetal acceleration?
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An apple falling to the ground is not an example of centripetal acceleration.
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