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Aleksandr [31]
3 years ago
13

Which atom is most likely to form a metallic bond ?

Physics
1 answer:
Nana76 [90]3 years ago
5 0

<u>Answer:</u>

"Aluminium (Al)" atom is most likely to form a metallic bond.

<u>Explanation:</u>

A kind of chemical bonding which generates between conduction electrons and positively charged metal ions from the attractive electrostatic force is understood as "metallic bonding". It may be defined as sharing free electrons among a positively charged ion structure.

A sheet of foil made from aluminum is constructed from metal bonds. Metallic relations are formed by strong forces of attraction. This property leads to low volatility, high melting and boiling points, and most metals are high density.

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Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
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Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

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