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Fittoniya [83]
2 years ago
5

What 2 elements on the periodic table are solid at room temperature?

Physics
2 answers:
shepuryov [24]2 years ago
8 0

Answer:

Eleven non-metals are gases at room temperature, including oxygen and chlorine. One non-metal, bromine, is a liquid at room temperature. The other non-metals are solids at room temperature, including carbon and sulfur.

Elan Coil [88]2 years ago
3 0

Answer:

All the elements are solids at 25 deg Celsius except mercury (Hg) and bromine (Br) that are liquids and hydrogen (H), nitrogen (N), oxygen (O), fluorine (F), chlorine (Cl), and the Noble gases in Family 18 that are gases.

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Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance
siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = k \frac{Q}{x_1^2} + k  \frac{Q}{x_2^2}

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = k  \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )

            E_ {total} = k \frac{Q}{x_1^2}   ( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

           

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = -k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = + k \frac{Q}{(d-x_1)^2}+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

             

5) Right side point

            E_ {total} = k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}

             E_ {total} = - k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0
2 years ago
A scientist is trying to determine the identity of an element. It is highly reactive to water and forms an ionic bond with chlor
wel

The answer is noble gas. Since noble gas are constant and unreactive. They can still shape compounds with other elements. 
Group 15 is also group 5A and Group 17 is also group 7A. Elements in these sets do not typically form ionic bonds; they are more on creating covalent bonds since they're non-metals. 
Therefore, that leaves us with B. from Group 1. They are metals (but Hydrogen) which respond violently with water, and they form ionic bonds, for they drop outer electrons easily.

8 0
3 years ago
Define si system of unit
Zepler [3.9K]
Good good good 7777777
8 0
3 years ago
An electric wall clock has a second hand 15 cm long. at the tip of his hand, what is the magnitude of the velocity?
Mila [183]

The velocity of the tip of the second hand is 0.0158 m/s

Explanation:

First of all, we need to calculate the angular velocity of the second hand.

We know that the second hand completes one full circle in

T = 60 seconds

Therefore, its angular velocity is:

\omega = \frac{2\pi}{T}=\frac{2\pi}{(60)}=0.105 rad/s

Now we can calculate the velocity of a point on the tip of the hand by using the formula

v=\omega r

where

\omega=0.105 rad/s is the angular velocity

r = 15 cm = 0.15 m is the radius of the circle (the distance of the point from the centre of rotation)

Substituting,

v=(0.105)(0.15)=0.0158 m/s

Learn more about angular motion here:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0
2 years ago
If the car goes exits a freeway and goes from 65<br> mph to 35 mph is it accelerating?
Zolol [24]

Answer:

No, the car is decelerating  

Explanation:

No the car is decelerating if it exits a freeway and goes from 65

mph to 35 mph since the change in velocity is negative.

change in velocity = final - initial

change in velocity  = 35 - 65

change in velocity = -30mph

Since the change in velocity is negative, hence the car is decelerating. Deceleration is a negative acceleration

8 0
3 years ago
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