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3 years ago

Which of the following explains why metallic bonding only occurs between

Physics

1 answer:

Y_Kistochka [10]3 years ago

3 0

Answer:

D. Metallic atoms have valence shells that are mostly empty, which

means these atoms are more likely to give up electrons and allow

them to move freely.

Explanation:

Metals usually contain very few electrons in their valence shells hence they easily give up these few valence electrons to yield metal cations.

In the metallic bond, metal cations are held together by electrostatic attraction between the metal ions and a sea of mobile electrons.

Since metals give up their electrons easily, it is very easy for them to participate in metallic bonding. They give up their electrons easily because their valence shells are mostly empty, metal valence shells usually contain only a few electrons.

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Does a 2000 mercury cougar have a timing belt

Katena32 [7]

No it does not have a timing belt

5 0

3 years ago

In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small spher

boyakko [2]

Answer:

L₀ = L_f , K_f < K₀

Explanation:

For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.

This means that the angular momentum before and after the collision changes.

Initial instant. Before the crash

L₀ = I₀ w₀

Final moment. Right after the crash

L_f = (I₀ + mr²) w

we treat the clay sphere as a point particle

how the angular momentum is conserved

L₀ = L_f

I₀ w₀ = (I₀ + mr²) w

w = $\frac{I_o}{I_o + m r^2}$ w₀

having the angular velocities we can calculate the kinetic energy

starting point. Before the crash

K₀ = ½ I₀ w₀²

final point. After the crash

K_f = ½ (I₀ + mr²) w²

sustitute

K_f = ½ (I₀ + mr²) ( $\frac{I_o}{I_o + m r^2}$ w₀)²

Kf = ½ $\frac{I_o^2}{ I_o + m r^2}$ w₀²

we look for the relationship between the kinetic energy

$\frac{K_f}{K_o}$ = $\frac{I_o}{I_o + m r^2}$

$\frac{K_f}{K_o } < 1$

K_f < K₀

we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision

6 0

3 years ago

Drag the correct labels to the images. Each label can be used more than once.

coldgirl [10]

Answer:

plato answer.

Explanation:

4 0

3 years ago

A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a

alisha [4.7K]

Answer:

a) x = v₀² sin 2θ / g

b) t_total = 2 v₀ sin θ / g

c) x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

sin θ = $v_{oy}$ / vo

cos θ = v₀ₓ / vo

v_{oy} = v_{o} sin θ

v₀ₓ = v₀ cos θ

v_{oy} = 13.5 sin 32 = 7.15 m / s

v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

v₀ₓ = x / t

x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

v_{y} = v_{oy} - gt

0 = v₀ sin θ - gt

t = v_{o} sin θ / g

we substitute

x = v₀ cos θ (2 v_{o} sin θ / g)

x = v₀² /g 2 cos θ sin θ

x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

v_{y} = v_{oy} - gt

v_{y} = 0

t = v_{oy} / g

t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

t_total = 2 t

t_total = 2 v₀ sin θ / g

c) we calculate

x = 13.5 2 sin (2 32) / 9.8

x = 16.7 m

4 0

3 years ago

If you act without reason or sound judgement, people will describe you as __________.

EastWind [94]

Answer:

irrational

Explanation:

8 0

3 years ago