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2 years ago

Hi:) why do metals have free electrons? anyone able to explain the conduction part as well? Thanks!

Physics

1 answer:

Karo-lina-s [1.5K]2 years ago

7 0

Metals have free electrons due to the bonding in metallic substances.

In a metal there are strong attractive forces between the nuclei and the valance electrons.

Positively charged metal nuclei form a lattice (a cube like structure) each metal atom provides one or more valance electrons <u>that are free to move throughout the lattice</u> The electrons are attracted to the positively charged nuclei but not one individual nuclei, this is called non-directional bonding since it occurs in all directions.

Now all metals are conductive becuase of the free to move (delocalised) electrons. Since the valance electrons are free to move throughout the lattice they are able to carry a charge. (Ionic solids cannot since the ionic solids form a tightly packed lattice with cations and anions which have no free moving electrons, electrons have to be able to move to carry a charge)

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Answer:

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Explanation:

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3 years ago

4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest

yarga [219]

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.

$v_{f}=v_{i}-gt$ (1)

Where:

v(i) is the initial velocity
v(f) is the final velocity
g is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

$0=30-9.81t$

Solve it for t:

$t=3.06\: s$

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.

$v_{f}^{2}=v_{i}^{2}-2gh$ (2)

$0=v_{i}^{2}-2gh$

We know that the initial velocity is 30 m/s.

$0=30^{2}-2gh$

Solving it for h we have:

$h=\frac{30^{2}}{2*9.81}$

$h=45.87 \: m$

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

$10=30-(9.81t)$

So, the time elapsed to get 10 m/s is:

$t_{upward}=2.04\: s$

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

$t_{upward}=2.04+t$

$t=1.02\: s$

This is the time when the ball has 10 m/s downward.

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

$t=2t_{upward}$

$t=2*3.06$

$t=6.12\: s$

The displacement will be zero after 6.12 s.

e) Here we need to find the time when v(f) is 15 m/s

$15=30-gt$

$t=\frac{15}{9.81}$

$t=1.53\: s$

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Learn more about vertical motion here:

brainly.com/question/13966860

I hope it helps you!

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2 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

SVEN [57.7K]

Answer:

a) Knowns, initial speed $v_{i}=13.0 m/s$ , final speed $v_{f}=0 m/s$ and gravity due it is a constant $g=9.8m/s^{2}$

b) The maximum high reached by the dolphin is $y_{max}=8.62 m$

c) Total time is $t=2.65s$

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at $0m/s$ speed.

b) Second, now that we know final speed we use $v_{f} =v_{i}-gt$ , as we clear for $t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s$ .

Then we use $y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2} =8.62m$

c)Third, finally we can use $y=v_{i}t-\frac{1}{2} gt^{2}$ , as we know $y=0m$ when the dolphin fall into the water again and $v_{i} =13.0m/s$ , then we have $0=(13m/s)t-\frac{1}{2} (9.8m/s^{2} )t^{2}$ is a quadratic form $0=t(13.0-4.9t)$ so we have $t_{i}=0s$ and $t_{f}=\frac{13}{4.90} =2.65s$