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2 years ago

A spherical electron cloud surrounding an atomic nucleus would best represent

Physics

1 answer:

podryga [215]2 years ago

6 0

Answer:

a S orbital

Explanation:

Atomic orbitals is the place where we are most likely to find at least one electron, this definition is based on the equation posed by Erwin Schrödinger.

It is said that each electron occupies an atomic orbital that is defined by a series of quantum numbers s, n, ml, ms. In any atom each orbital can contain two electrons. It is possible that thanks to the function of the orbitals, the appearance that atoms can have is that of a diffuse cloud.

The orbitals s (l = 0) have a spherical shape. The extent of this orbital depends on the value of the main quantum number, so a 3s orbital has the same shape but is larger than a 2s orbital.

The orbitals p (l = 1) are formed by two identical lobes that project along an axis. The junction zone of both lobes coincides with the atomic nucleus. There are three orbitals p (m = -1, m = 0 and m = + 1) in the same way, which differ only in their orientation along the x, y or z axes.

The orbitals d (l = 2) are also formed by lobes. There are five types of d orbitals (corresponding to m = -2, -1, 0, 1, 2)

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An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

4 0

2 years ago

Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m

OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

$E_m = \frac{1}{2} m\omega_1^2A_1^2$

The energy of the system having mass 2m is,

$E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2$

For the two expressions mentioned above remember that the variables mean

m = mass

$\omega =$ Angular velocity

A = Amplitude

The energies of the two system are same then,

$E_m = E_{2m}$

$\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2$

$\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}$

Remember that

$k = m\omega^2 \rightarrow \omega^2 = k/m$

Replacing this value we have then

$\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}$

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}$

But the value of the mass was previously given, then

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}$

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}$

$\frac{A_1}{A_2} = 1$

Therefore the ratio of the oscillation amplitudes it is the same.

5 0

2 years ago

The rear wheel on a clown’s bicycle has twice the radius of the front wheel. (a) When the bicycle is moving, is the linear speed

8_murik_8 [283]

Answer:

a). same as

b). less than

Explanation:

a). When a bicycle is moving, the linear speed at the top of the rear wheel is same as the linear speed at the top of the front wheel. Since the clown's bicycle is a rigid body, both the wheels that is the front wheel and the rear wheel will move with the same linear speed.

b). Since we know that angular speed varies inversely to the radius of the wheel.

That is ω = 1 / r

Since the rear wheel has twice the radius of that of the front wheel, therefore real wheel will have less angular speed than the front wheel.

Therefore, the angular speed of the rear wheel is less than the angular speed of the front wheel.

7 0

2 years ago

The colored lines in the figure represent paths taken by different people walking around in a city. Assume that each city block

jarptica [38.1K]

Answer: 592.37m

Explanation:

Person D is the blue line.

The total displacement is equal to the difference between the final position and the initial position, if the initial position is (0,0) we have that he first goes down two blocks, then right 6 blocks. then up 4 blocks, then left 1 block.

Now i will considerate that the positive x-axis is to the right and the positive y-axis is upwards.

Then the new position will be, if B is a block:

P =(6*B - 1*B, -2*B + 4*B) = (5*B, 2*B)

And we know that B = 110m

P = (550m, 220m)

Now, then the displacement will be equal to the magnitude of our vector, (because the difference between P and the initial position is equal to P, as the initial position is (0,0)) this is:

P = √(550^2 + 220^2) = 592.37m

4 0

2 years ago

Centripetal force Fc acts on a car going around a curve. If the speed of the car were twice as great, the magnitude of the centr

kondaur [170]

Answer:

We need 4 times more force to keep the car in circular motion if the velocity gets double.

Explanation:

Lets take the mass of the car = m

The radius of the arc = r

$F=\frac{m\times v^2}{r}$