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3 years ago

List four methods that can be used to separate mixtures and give an example of each method?

1 answer:

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(1) Filtration - example is separating saltwater from sand particles.
(2) Distillation - example is apple cider vinegar. Distillation collects the liquid that we want, which is the vinegar and leave out other heavier suspended particles.
(3) Centrifuges - example is separating the blood components. They separate the plasma and erythrocytes.
(4) Magnetism - example is separating metals from non-metals in electromagnetic cranes, that differentiate between metal and heap of junks.

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Describe an experiment to determine how the frequency of a vibrating string depends on the length of the string

Ksivusya [100]

Answer:

For a vibrating string, the fundamental frequency depends on the string's length, its tension, and its mass per unit length. ... The fundamental frequency of a vibrating string is inversely proportional to its length.

Explanation:

Sounds of a single pure frequency are produced only by tuning forks and electronic devices called oscillators; most sounds are a mixture of tones of different frequencies and amplitudes. The tones produced by musical instruments have one important characteristic in common: they are periodic, that is, the vibrations occur in repeating patterns. The oscilloscope trace of a trumpet's sound shows such a pattern. For most non-musical sounds, such as those of a bursting balloon or a person coughing, an oscilloscope trace would show a jagged, irregular pattern, indicating a jumble of frequencies and amplitudes.

A column of air, as that in a trumpet, and a piano string both have a fundamental frequency—the frequency at which they vibrate most readily when set in motion. For a vibrating column of air, that frequency is determined principally by the length of the column. (The trumpet's valves are used to change the effective length of the column.) For a vibrating string, the fundamental frequency depends on the string's length, its tension, and its mass per unit length.

In addition to its fundamental frequency, a string or vibrating column of air also produces overtones with frequencies that are whole-number multiples of the fundamental frequency. It is the number of overtones produced and their relative strength that gives a musical tone from a given source its distinctive quality, or timbre. The addition of further overtones would produce a complicated pattern, such as that of the oscilloscope trace of the trumpet's sound.

How the fundamental frequency of a vibrating string depends on the string's length, tension, and mass per unit length is described by three laws:

1. The fundamental frequency of a vibrating string is inversely proportional to its length.

Reducing the length of a vibrating string by one-half will double its frequency, raising the pitch by one octave, if the tension remains the same.

2. The fundamental frequency of a vibrating string is directly proportional to the square root of the tension.

Increasing the tension of a vibrating string raises the frequency; if the tension is made four times as great, the frequency is doubled, and the pitch is raised by one octave.

3. The fundamental frequency of a vibrating string is inversely proportional to the square root of the mass per unit length.

This means that of two strings of the same material and with the same length and tension, the thicker string has the lower fundamental frequency. If the mass per unit length of one string is four times that of the other, the thicker string has a fundamental frequency one-half that of the thinner string and produces a tone one octave lower.

7 0

3 years ago

A roulette wheel with a 1.0m radius reaches a maximum angular speed of 18 rad/s before it stops 35 revolutions ( 220 rad ) after

Alex17521 [72]

Max ang. speed(u) = 18 rad/s
final ang. speed(v) = 0
ang. displacement(s) = 220 rad

ang. acceleration = (v^2 - u^2)/2s = -18^2 / 2*220 = -0.7364 rad/s^2

v = u +at
0 = 18 - 0.7364t
t = 18/0.7364
t = 24.44 seconds

4 0

3 years ago

4. I drop a pufferfish of mass 5 kg from a height of 5.5 m onto an upright spring of total length 0.5 m and spring constant 3000

KatRina [158]

Answer:

a) 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b) at the height of 0.484 m height , the pufferfish will eventually come to rest.

c) There exists two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy = 23.72J

Spring potential energy = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

Now by using the conservation of energy; we realize that :

Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

$mgh = mgx + \frac{1}{2}K(l-x)^2$

Replacing our given values into the above equation; we have :

$(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2$

269.5 = 47.5 x + 1500(0.5 -x )²

269.5 = 47.5 x + 1500(0.25 - x²)

269.5 = 47.5 x + 375 - 1500 x²

269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m : so this is the compression in the spring

Now; to determine the height the pufferfish gets to before it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

There exists two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy

$=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J$

8 0

3 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago

Fill in the blanks with suitable pronouns.

DaniilM [7]

1)Baby bird cannot fly. Their mother has to feed \bf\underline{them}them

2. Vijay likes riding my bicycle. I sometimes lend \bf\underline{it}it to \bf\underline{him}him

3. Sooraj and I are brothers. \bf\underline{we\:both}weboth share the same bedroom.

4. Ravina isn't well. Dad is taking to the doctor. (No personal pronouns required)

<h3>5. My sister is a teacher. \bf\underline{She}She teaches Maths.</h3>

7 0

3 years ago