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Kay [80]
3 years ago
10

Which statement about electrons and atomic orbitals is not true?

Physics
2 answers:
Lubov Fominskaja [6]3 years ago
5 0

An electron cloud represents all the orbitals in an atom.

joja [24]3 years ago
3 0

Answer:

An electron cloud represents all the orbitals in an atom.

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ENERGY SAVERS RACE, BRAIN BURNER. This question is about solar cars at the Chuck Norris Institute of Technology, CNIT, in Ocala,
Hunter-Best [27]

Answer:

a) d = 6.0 m

Explanation:

Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as

d = \frac{(v_f + v_i)}{2} \times t

here we know that

v_f = 10 m/s

v_i = 0

t = 1.2 s

now we will have

d = \frac{(10 + 0)}{2}\times 1.2

d = 5 \times 1.2

d = 6.0 m

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3 years ago
What is happening in the diagram?
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An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, u=4.52\times 10^5\ m/s

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

v^2=u^2+2as........(1)

a is the acceleration, a=\dfrac{F}{m}

We know that electric force, F = qE

a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

or

v=6.45\times 10^5\ m/s

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3 0
3 years ago
A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s. Assume in this problem that air drag i
stiks02 [169]

Answer:

P.E = 0.068 J = 68 mJ

Explanation:

First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = -9.8 m/s²  (negative sign due to upward motion)

h = height attained by the ball toy = ?

Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)

Vi = Initial Velocity = 3 m/s

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2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²

h = (9 m²/s²)/(19.6 m/s²)

h = 0.46 m

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P.E = mgh

P.E = (0.015 kg)(9.8 m/s²)(0.46 m)

<u>P.E = 0.068 J = 68 mJ</u>

4 0
3 years ago
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