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3 years ago

Why are the parts of an atom that electrons occupy called electron clouds?

1 answer:

8 0

Because it's literally impossible to tell exactly where something that size is
located at any particular time.

And that's NOT because it's so small that we can't see it. It's because any
material object behaves as if it's made of waves, and the smaller the object is,
the more the size of its waves get to be like the same size as the object.
When you get down to things the size of subatomic particles, it doesn't make
sense any more to try and talk about where the particle actually "is", and we only
talk about the waves that define it, and how the waves all combine to become a
cloud of <em><u>probability</u></em> of where the particle is.

I know it sounds weird. But that's the way it is. Sorry.

You might be interested in

Suppose you exert a 400-N force on a wall but the wall does not move. The work you are doing on the wall is : unknown, because t

Ilya [14]

Answer:

W = 0 :The work done on the wall is zero,because the wall is not moving

Explanation:

Work theory

Work is the product of a force applied to a body and the displacement of the body in the direction of this force.

W= F*d Formula (1)

W: Work (Joules) (J)

F: force applied (N)

d=displacement of the body (m)

The work is positive (W+) if the force goes in the same direction of movement.

The work is negative (W-)if the force goes in the opposite direction to the movement

Data

F= 400-N

d= 0

Problem development

We apply formula (1) to calculate the work done on the wall:

W= 400*0

W=0

6 0

3 years ago

A/ Can someone help me, how can the third law of motion (Newton) help solve a problem / issue. be specific

mezya [45]

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sorry this is by accident, but i can't delete it

6 0

3 years ago

Name the type of simple machine: A ramp on the back of a moving van.

Mrac [35]

C) an inclined plane

6 0

2 years ago

Read 2 more answers

Pigments are always made of natural materials

Kobotan [32]

It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway, the answer to the given statement is FALSE. Pigments are NOT always made of natural materials. Pigments can either be man-made or the one that occurred naturally as what you see in our environment. Hope this helps.

5 0

2 years ago

Two wires of the same material and having the same volume, are fixed

Setler79 [48]

Answer:

48 kg

Explanation:

Given that the two wires are of same material, so their value of young's modulus will be same

Assuming that the wires are cylindrical in shape

As radius of the first wire is half that of the second wire and therefore the area of cross-section of the first wire will be one-fourth of the second wire( ∵ wire is cylindrical, the cross-sectional part will be circle and the area of the circle = π × r² )

As the volume is same for both wires

∴ π × ( $r_{1}$ )² × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$

Here

$r_{1}$ is the radius of the first wire

$r_{2}$ is the radius of the second wire

$l_{1}$ is the length of the first wire

$l_{2}$ is the length of the second wire

⇒ π × (( $r_{2}$ )² ÷ 4) × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$ (∵ radius of first wire is half that of the second wire)

By cancelling the same terms on both sides

we get

$l_{1}$ = 4 × $l_{2}$

⇒ Length of first wire will be four times of the length of second wire

<h3>Strain is defined as the elongation per unit length</h3>

Strain in first wire = ΔL ÷ $l_{1}$ = ΔL ÷ (4 × $l_{2}$ )

where ΔL is the elongation of the wire which in this case is same in both wires

Strain in second wire = ΔL ÷ $l_{2}$

∴ Strain in second wire is four times of strain in first wire

<h3>Stress = F ÷ A</h3>

where F is the force perpendicular to the cross-sectional area

A is the area of cross-section

Force in first wire = $m_{1}$ × g

where $m_{1}$ is the mass hanged to the first wire

g is the acceleration due to gravity

Force in second wire = $m_{2}$ × g

where $m_{2}$ is the mass hanged to the second wire

g is the acceleration due to gravity

Let $A_{1}$ be the cross-sectional area of first wire

$A_{2}$ be the cross-sectional area of second wire

$A_{2}$ = 4 × $A_{1}$ (∵ cross=sectional area of the wire = π × (radius of the wire)² )

Stress in first wire = ( $m_{1}$ × g) ÷ ( $A_{1}$ )

Stress in second wire = ( $m_{2}$ × g) ÷ ( $A_{2}$ ) = ( $m_{2}$ × g) ÷ (4 × $A_{1}$ )

<h3>Young's modulus is defined as Stress per unit strain</h3>

As Young's modulus is same for both wires, Stress per unit strain must be same for both wires

Stress per unit strain of first wire = (( $m_{1}$ × g) ÷ ( $A_{1}$ )) ÷ (ΔL ÷ (4 × $l_{2}$ ))

Stress per unit strain of second wire = (( $m_{2}$ × g) ÷ (4 × $A_{1}$ )) ÷ (ΔL ÷ $l_{2}$ )

By equating them we get

$m_{2}$ = 16 × $m_{1}$

⇒ $m_{2}$ = 16 × 3 = 48 kg

∴ $m_{2}$ = 48 kg

5 0

3 years ago