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Lisa [10]
3 years ago
12

Elaborate on the classification of aluminum. A) Aluminum is a noble gas and has a full valence electron shell, is chemically non

-reactive, colorless, odorless, and is a gas at room temperature. B) Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution. C) Aluminum is a nonmetal and is dull, brittle, poor conductor of heat and electricity, forms acidic oxides, and forms anions in aqueous solution. D) Aluminum is a metalloid and has properties of both metals and nonmetals - it appears lustrous, but is brittle and a poor conductor of heat and electricity.
Physics
2 answers:
guajiro [1.7K]3 years ago
5 0

Elaborate on the classification of aluminum.

                                         *correct answer*


B) Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.

                                                *yw*

scZoUnD [109]3 years ago
4 0
The correct option is (B) <span>Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.

Since Aluminium is in group 13, and all the elements in group 13 are either metals or metalloids(Boron). Hence we are left with option (B) and (D). Boron is the only metalloid in group 13 and aluminium is a metal(not a metalloid); therefore, we are left with only one option which is Option (B). And Aluminium is </span>shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.<span>
</span><span>
</span>
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An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-
maw [93]

Answer:

  1. Power requirement <u>P</u> for the banner is found to be  30.62 W
  2. Power requirement <u>P</u> for the solid flat plate is found to be 653.225 W
  3. Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The <em>Cd </em>for banner was given, whereas the <em>Cd </em>for a flat plate is generally found to be around <em><u>1.28</u></em><em> </em>which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
  4. Power requirement <u>P</u> for the smooth spherical balloon was found to be 40.08 W

Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

  1. v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
  2. The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
  3. The power can be determined this equation: P = F . v, where F represents <em>the drag-force</em> that we will need to determine and v represents the<em> velocity of the airplane</em>
  4. The equation to determine drag-force is: F = 1/2 * d *  C_d * A

In the drag-force equation Cd represents the c<em>o-efficient of drag</em> and A represents the <em>frontal area of the banner/plate/balloon (the object being towed)</em>

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

<u>Part a)</u> We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = <em>1/2 * 0.06* 1.225 * 20</em> = 0.735 N. Now to find the power P we will use P = F . v i.e.<em> 0.735 * 41.66</em> = <u><em>30.62 W</em></u>

<em></em>

<u>Part b) </u>For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = <em>1/2 * 1.28 * 1.225 * 20</em> = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = <u><em>653.225 W</em></u>

<u><em></em></u>

<u>Part c)</u> The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

<u>Part d)</u> The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : <em>pi* r^2 (r = d / 2).</em> Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = <u><em>40.08 W</em></u>

4 0
3 years ago
You have a 35X objective lens in place, and the numerical aperture of the objective lens is 0.75. The numerical aperture of the
olasank [31]

Answer:

350x

Explanation:

In a microscope the objective has higher magnification than the eyepiece so, this is a microscope

The magnification of a microscope is given by the product of the magnifications of the eyepiece and and the objective.

Objective lens magnification = 35x =m_o

Eyepiece magnification = 10x =m_e

Total magnification

M=m_o\times m_e\\\Rightarrow M=35\times 10\\\Rightarrow M=350

Total magnification is 350x

8 0
3 years ago
What is the mean of 22, 72, 79, 72, and 70
Nataly_w [17]

Answer:

63

Explanation:

You first have to add all the numbers together.

22+72+79+72+70 = 315

You divide the total by the amount of numbers (5)

315/5 = 63

The mean is 63

7 0
3 years ago
What type of collision is being described in each statement.
Ostrovityanka [42]

Answer:

inelastic, since the girl moves in the same direction as the thrown ball

Explanation:

yess this ok

UwU

7 0
2 years ago
A wave with a wavelength of 0.1 m moves with a speed of 0.5 m/s. What is the frequency of the wave?
kondaur [170]
Frequency= speed/ wavelength
=0.5m/s divided 0.1.m
=5.0 Hz

The answer would be letter D.
3 0
3 years ago
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