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Nikolay [14]
2 years ago
6

2) Imagine that a loudspeaker is producing quiet note with a low pitch

Chemistry
1 answer:
makvit [3.9K]2 years ago
3 0

Answer:

Higher pitched sounds produce waves which are closer together than for lower pitched sounds. (Think of the slinky - if you produce waves rapidly they will travel quite close to each other, thus demonstrating a higher pitched sound.)

The pitch of a note will depend on a number of factors. One of these is the size of the vibrating object. On a glockenspiel or xylophone the high notes are made by the smaller bars. A smaller triangle or cymbal will make a relatively higher pitch note. On a stringed instrument such as a guitar or violin a thinner string will generally make a higher note, but also shortening the string by stopping it with the finger will produce a similar effect. On a set of pan pipes or a church organ it is the shorter pipes which make the higher notes when the air inside them vibrates.

Another factor which produces higher pitched notes is the tension within the vibrating object. A guitar string can be tuned to a higher pitch by adjusting the string tensioner. An elastic band can be stretched tighter and a drum skin can be tensioned to increase the pitch of the sound it produces.

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A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and
lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

F_{solution} = mass of solvent + mass of oil (i.e A+C)

<u>Feed Phase:</u>

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

 = 25kg

Mass ratio of oil to solution Y_{F} =\frac{Mass C}{Mass (A+C)}

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane  in feed = 25 ×  0.9 =22.5kg + 2.5 =25kg

therefore  Y_{F} = \frac{2.5}{25}

= 0.1

mass ratio of solid to solution Y_{A}  =  \frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}

=3

<u>Solvent Phase:</u>

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

<u>Underflow:</u>

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = \frac{mass B}{mass(A+C)}

solution in underflow E₁ = Mass (A+C)

<u>Overflow:</u>

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75=  E₁ × N₁

75 =  E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

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