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3 years ago

Electrons are modeled using a "cloud" around the nucleus because they

Physics

2 answers:

LenaWriter [7]3 years ago

7 0

Explanation:

It is known that an atom consists of three sub-atomic particles, that is, protons, neutrons, and electrons.

Protons and neutrons are located inside the nucleus of the atom whereas electrons revolve around the nucleus of the atom forming a cloud of electrons.

These electrons occupy orbits of their respective energy levels.

Therefore, we can conclude that electrons are modeled using a "cloud" around the nucleus because they revolve around the nucleus.

taurus [48]3 years ago

4 0

An electron cloud is a visual model of the most probable locations of electrons in a atom. The cloud is denser where you will probably find a electron.

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A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp

raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination $\theta =11.1^{\circ}$

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction $\mu _k=0.39$

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using $v^2-u^2=2as$

Final velocity v=0

$0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s$

$s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}$

s=0.68 m

5 0

3 years ago

What is projectile motion

MakcuM [25]

<h2><u>Projectile</u><u> </u><u>motion</u><u>:</u></h2>

<em>If</em><em> </em><em>an</em><em> </em><em>object is given an initial velocity</em><em> </em><em>in any direction and then allowed</em><em> </em><em>to travel freely under gravity</em><em>, </em><em>it</em><em> </em><em>is</em><em> </em><em>called a projectile motion</em><em>. </em>

It is basically 3 types.

horizontally projectile motion
oblique projectile motion
included plane projectile motion

7 0

3 years ago

Read 2 more answers

A friend tosses a baseball out of his second floor window with initial velocity of 4.3m/s(42degrees below the horizontal). The b

schepotkina [342]

<span>b)Determine your horisontal distance from window (ans. 1.5 m)
c)Calc the speed of ball as you catch it (ans: 8.2m/s)

I dont get what 42 m below the horizontal is, can someone give me direction on how to do this? </span>

7 0

3 years ago

Read 2 more answers

A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto

sesenic [268]

<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float= $l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water= $10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$ cm

Hence, the depth of barge float=3 cm

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4 0

4 years ago

Read 2 more answers

A car moves along a curved road of diameter 2 km. If the maximum velocity for safe driving on this path is 30 m/s, at what angle

Leto [7]

The maximum velocity in a banked road, ignoring friction, is given by;

v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.

Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°

Therefore, the road has been banked at 5.24°.

4 0

4 years ago