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3 years ago

Match these terms with the correct examples.

2 answers:

4 0

1. liquid solution to a. oceans
2. gaseous solution to b. clouds
Not sure about 3 and 4.
3 might be oxygen but I think that's 5. element.

Hope this helps, not sure about water and air though.

kogti [31]3 years ago

3 0

<h3>Answer:</h3>

1) Liquid solution: Oceans

2) Gaseous Solution: Air

3) Compound: Water

4) Colloid: Cloud

5) Element: Oxygen

<h3>Explanation:</h3>

Solutions are those mixtures which have uniform composition. The size of solute is around 1 Angstrom hence, the particles of solutes can not be seen and the interactions make the overall solution transparent.

Compounds are those substances which are formed chemically by the reaction of two different elements like,

O₂ + 2 H₂ → 2 H₂O

Colloids are also mixtures but these are intermediate between true solutions and suspensions. Also, the solute size is greater than that of solute size in solutions and is about 10 Angstrom. Examples include Clouds, Fog, Moist, Aerosol sprays e.t.c.

Elements are those substances which either are made up of single atoms like noble gases or are molecules made up of same elements like O₂, H₂, Cl₂, N₂ e.t.c.

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2 years ago

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3 years ago

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3 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

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2 years ago