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2 years ago

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The quantum state of a particle can be specified by giving a complete set of quantum numbers (n,l, ml,ms). How many different qu

antum states are possible if the principal quantum number is n = 5?

1 answer:

GarryVolchara [31]2 years ago

7 0

The total number of different quantum states are possible if the principal quantum number is n = 5 are 50.

<h3>What is quantum number?</h3>

Quantum number is the set of numbers that describe the position and energy of the electron in an atom.

The allowed orbital quantum numbers l and ml are

0 ............1

-1, 0, +1 .................3

-2, -1, 0, 1, 2 ...................5

-3, -2, -1, 0,1 ,2,3...................7

-4,-3,-2,-1,-,1,2,3,4....................9

Total = 2 x (1+3+5+7+9)

= 50

Thus, 50 different quantum states are possible.

Learn more about quantum number.

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A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.

Norma-Jean [14]

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

$0=m_{bullet}*v_{bullet} + m_{rifle}*v_{rifle} \\$

assume the bullet goes to right side and the gravitational acceleration =10 $ms^{-2}$

so now the weight of the rifle= $\frac{25}{10}$

$0=m_{bullet}*v_{bullet} + m_{rifle}*v_{rifle} \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}$

this is a negative velocity to the right side. that means the rifle recoils to the left side

3 0

3 years ago

A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m

lbvjy [14]

The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the <em>mass </em>of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the<em> lab cart </em>

$v_{2}_{i}$ : is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the<em> lab cart </em>

$v_{2}_{f}$ : is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

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I hope it helps you!

4 0

3 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

3 years ago

All rules to find work in changing force in physics

saw5 [17]

Answer:

FORCE - rate of change of momentum, ie its changing velocity [change in velocity is of more concern] NEWTON

WORK - product of force and displacement, ie [velocity may be constant or variable but change in position with certain force is of more concern] JOULES

I hope you understood from this..

7 0

2 years ago

The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N

guajiro [1.7K]

Answer:

$\frac{r_1}{r_2}=6.9$

Explanation:

According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:

$P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}$

where,

F₁ = Load lifted by output plunger = 2100 N

F₂ = Force applied on input piston = 44 N

r₁ = radius of output plunger

r₂ = radius of input piston

Therefore,

$\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9$

4 0

3 years ago