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3 years ago

Why the substance on the periodic talbe classified together

Physics

2 answers:

denpristay [2]3 years ago

6 0

you mean table*, but idk

Thepotemich [5.8K]3 years ago

5 0

Hey there!

Here is your answer for this question!

Answer : The elements in a group look and behave similary because they both have the same number of Electrons as their outermost protective shell. Also often said as the face they show to the world around us. Group number 18 elements as on the far side to the right of the Periodic Table, for example, you have completely full outer shells and hardly participate in the chemical reactions you might see in a substance.

The groups of the periodic table of groups 1,2, 11,15,16,17 and 18 are down below!

Group #1. Alkali Metals.

Group #2. Alkaline Earth Metals.

Group #11. Coinage Metals (it is not an IUPAC approved as the name!)

Group #15. Pnictogens (also not used as an IUPAC approved name!)

Group #16. Chalcogens.

Group #17. Halogens.

Group #18. Noble Gases.

I hope this helps you! Let me know if you need anything else! ☺

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Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe

NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B ) Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

= 71.42 m

6 0

2 years ago

Read 2 more answers

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

a. The frequency with which the waves strike the hill is 242.61 Hz

b. The frequency of the reflected sound wave is 254.23 Hz

c. The beat frequency produced by the direct and reflected sound is

11.62 Hz

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

2 years ago

A mine elevator is supported by a single steel cable 0.0125 m in diameter. The total mass of the elevator cage and occupants is

Sonja [21]

Answer:

0.8895m

Explanation:

Cable diameter = 0.0125m

Mass of elevator = 6450kg

Young Modulus(E) = 2.11*10¹¹N/m

∇l (change in length) =

L = 362m

A = Πr², but r = d / 2 = 0.0125 / 2 = 0.00625m

A = 3.142 * (0.00625)² = 1.227*10^-4m²

Young Modulus (E) = Tensile stress / Tensile strain

E = (F / A) / ∇l / L

F = mg = 6450 * 9.8 = 63210N

2.11*10¹¹ = (63210 / 1.22*10^-4) / (∇l / 362)

2.11*10¹¹ = 5.18*10⁸ / (∇l / 362)

2.11*10¹¹ = (5.18*10⁸ * 362) / ∇l

2.11*10¹¹ = 1.875*10¹¹ / ∇l

∇l = 1.875*10¹¹ / 2.11*10¹¹

∇l = 0.8895m

The change in length is 0.8895m

8 0

2 years ago

Besides, the discovery that moons orbit Jupiter, what other discovery made by Galileo Galilei, with an early telescope, proved t

Thepotemich [5.8K]

Answer:

He made great advancements in developing a logical way to know more about the universe and celestial entities inside the space. And this theory is termed to be heliocentric in nature.

Explanation:

In early times most of the people believed that our planet Earth is the center of the universe or the solar system and rest of the celestial entities move around it in a given path, so, it confused the well known scientist named as Galileo Galilei. As, he observed the different dark patches or shadow like textures on the face of the Sun.

While, it is more obvious to known that any object having multiple small shadows means that it is present inside such a region that all of the celestial entities move or orbit around it in a given way.So, he concluded that planet Earth itself move around the red Giant in a given way rather then being the center of the universe.

5 0

2 years ago

6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d

kipiarov [429]

Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

Vθ² = 15.75

Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

7 0

3 years ago

Why the substance on the periodic talbe classified together​

Why the substance on the periodic talbe classified together