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Bas_tet [7]
3 years ago
5

When two atoms share electrons the bond is ______

Physics
2 answers:
kipiarov [429]3 years ago
8 0
Hey there!

Here is your answer:

<u><em>The proper answer to this question is option A (or 1) "covalent".</em></u>

Reason:

<u><em> A covalent is a chemical bond that was made from to atoms which was shared by electrons and they stuck together to form a covalent.</em></u>

<em>Therefore the answer is 1.</em>

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit
Ksju [112]3 years ago
3 0
I think it is equal  for your question

You might be interested in
The temperature of the surface of the Sun is 5500°C.
natita [175]

Answer:

a. the average kinetic energy of hydrogen atoms is  1.20 × 10^-19J

b. the average kinetic energy of helium atoms is 1.24 × 10^-17J

Explanation:

The computation is shown below;

As we know that

Kinetic energy = 3 ÷ 2 kT

where,

K = Boltzmann constant

And, T = Temperature

a. Now the temperature in kelvin is

T = (5,500 × (°C ÷ K) + 273.15 K)

= 5773.15 K

As

Kinetic energy = 3 ÷ 2 kT

So now 1.38 × 10^-23 J/K for K would be substituted and 5773.15 K for Temperature T

Now Kinetic energy is

= 3 ÷ 2 (1.38 × 10^-23 J/K) ( 5773.15 K)

= 1.20 × 10^-19J

hence, the average kinetic energy of hydrogen atoms is  1.20 × 10^-19J

b. As

Kinetic energy = 3 ÷ 2 kT

now 1.38 × 10^-23 J/K for K would be substituted and 6 × 10^5K for Temperature T

Now Kinetic energy is

= 3 ÷ 2 (1.38 × 10^-23 J/K) (6 × 10^5K )

= 1.24 × 10^-17J

hence, the average kinetic energy of helium atoms is 1.24 × 10^-17J

8 0
3 years ago
1. The thermal convection that drives plate motion is caused by
noname [10]
Subduction I believe
6 0
3 years ago
A rubber ball is dropped from a height of 8m. After strikingthe floor, the ball bounces to a height of 5m. a. If the ball had bo
kifflom [539]

Answer:

a) This means the collision between the ball and the floor is elastic.

b) This points to a perfectly inelastic collision between the ball and the floor as they stick together after collision

c) Check Explanation.

Explanation:

Collision of bodies are analysed according to whether both momentum and kinetic energy of the system is conserved, that is, if these two quantities before collision are equal to their values after collision.

In all types of collisions, momentum is usually conserved, but kinetic energy is conserved only in an elastic collision.

A ball dropped from a height of 8 m bounces up back to a height of 5 m.

a. If the ball had bounced to a height of 8m, how would you describe the collision between the ball and the floor?

The ball not bouncing back to a height of 8 m shows energy loss at some point in the total motion of the ball (most likely at the collision). If kinetic energy was conserved, the ball would bounce back up to the height at which it fell from (8 m) after the collision with the floor.

b. If the ball had not bounced at all, how would you describe the collision between the ball and the floor?

If the ball had not bounced at all, this means it lost all of its kinetic energy to the floor, and this points to a perfectly inelastic collision between the ball and the floor as they stick together after collision.

c. What happened to the energy lost by the ball during thecollision?

The energy lost during the collision is converted to another form, most likely responsible for some deformation on the ball & a minute deformation on the floor, converted to some form of heat as a result of the collision or into sound energy, usually, it's a combination of all This!

Hope this Helps!!!

5 0
4 years ago
A cylinder with moment of inertia I1 rotates with angular speed ω0 about a frictionless vertical axle. A second cylinder, with m
MAXImum [283]

Answer:

Part(a): The final angular velocity is \omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}

Part(b): The ratio of the rotational energies is \dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})},showing the the energy of th system will decrease.

Explanation:

Part(a):

If 'I' be the moment of inertia of an object and '\omega' be its angular velocity then the angular momentum 'L' of the object can  be written as

L = I \omega

If 'I_{1}' and 'I_{2}' be the moment of inertia of the two cylinders and '\omega_{1}' and '\omega_{2}' be the initial angular velocity of the cylinders and '\omega_{1}{'}' and '\omega_{2}^'}' be their respective final angular velocity, then from conservation of angular momentum,

I_{1} \omega_{1} + I_{2} \omega_{2} = I_{1} \omega_{1}^{'} + I_{2} \omega_{2}^{'}

Given, \omega_{1} = \omega_{i},~\omega_{2} = 0,~\omega_{1}^{'} = \omega_{2}^{'} = \omega_{f}. From the above expression

&& I_{1} \omega_{i} = (I_{1} + I_{2}) \omega_{f}\\&or,& \omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}

Part(b):

Initial kinetic energy  

K_{i} = \dfrac{1}{2} I_{1} \omega_{i}^{2}

and Final kinetic energy

K_{f} = \dfrac{1}{2}(I_{1} + I_{2}) \omega_{f}^{2}

Substituting the value of \omega_{f},

&& K_{f} = \dfrac{1}{2}(I_{1} + I_{2})\dfrac{I_{1}^{2}\omega_{i}^{2}}{(I_{1} + I_{2})^{2}} = \dfrac{1}{(I_{1} + I_{2})} \dfrac{1}{2}I_{1}\omega_{i}^{2} = \dfrac{1}{(I_{1} + I_{2})} K_{i}\\&\dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})}

The above expression shows that the ebergy of the system will decrease.

7 0
3 years ago
What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between t
vlabodo [156]

Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force  between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

6 0
3 years ago
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