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MrMuchimi
4 years ago
11

What is the element with the lowest electronegativity value?

Physics
1 answer:
Lerok [7]4 years ago
7 0
<h2>Answer: Francium </h2>

Let's start by explaining that electronegativity is a term coined by Linus Pauling and is determined by the <em>ability of an atom of a certain element to attract electrons when chemically combined with another atom. </em>

So, the more electronegative an element is, the more electrons it will attract.

It should be noted that this value can not be measured directly by experiments, but it can be determined indirectly by means of calculations from other atomic or molecular properties of the element. That is why the scale created by Pauling is an arbitrary scale, where the maximum value of electronegativity is 4, assigned to Fluorine (F) and the <u>lowest is 0.7, assigned to Francium (Fr).</u>

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1. Which type of wire would be a better conductor of an electrical current?
Lelechka [254]

Answer:

1. a

2. b

3. b

Explanation:

1.

Resistance is the property of a conductor to offer resistance to the flow of current. The lower the resistance better is  the conductivity of wire.

We know that the resistance of a wire depends on several factor which are inter-connected by an equation as:

R=\rho.\frac{l}{a}

where:

R = resistance of the wire

l = length of the wire

a= cross sectional area of the wire  

from the above relation we observe that

R\propto l

R\propto \frac{1}{a}

  • Also when the temperature of the wire is significantly high then the lattice vibration cause obstruction in the path of the flowing charges and reduce the current flow.

2.

As the collision between the electrons and protons increases the speed of the flow of charges will decrease because the opposite charges attract each other and as we know that electrical current is the rate of flow of charge.

3.

Heating up of wire due to sunlight will cause lattice vibration in the conductor and will obstruct the movement of the charges which build up electric current, hence increasing the resistance of conductivity.

7 0
3 years ago
A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp
Arte-miy333 [17]

Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m,   m2 = 4m,    v1=vf_p,    v2 = vf_alpha

The conservation momentum states that:

pi = pf      

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

      m1 =  m, m2 = 4m, q1 = e, q2 = 2e

      and   v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 =  k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

3 0
3 years ago
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