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Leviafan [203]
3 years ago
5

Consider a system of two particles with spin quantum numbers 1andS2respectivelyand with no orbital angular momentum. We have see

n in class that this state can be described equally well in two different bases:
Physics
1 answer:
o-na [289]3 years ago
5 0

Answer:

The Clebsh-Gordan Coefficients to relate the coupled and uncoupled bases.

Explanation:

In quantum mechanics, two different sources of angular momentum eigenstates, show their widening through the Clebsch–Gordan (CG) coefficients, first in an uncoupled product bases.

- The uncoupled basis writes the state as eigenstates of the z-components of    the two particles:

                               IS1+S2ms1ms2)

- The coupled basis writes the state as eigenstates of the two particles:

                                IS1S2Stotms)

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What is the series equivalent of two 1000 W resistors in series?
aleksandrvk [35]
The equivalent resistance when two resistors are connected in series is
the sum of their individual resistances.

The marking on the resistor that says "1000 W" is the rating that tells
how much power the resistor can safely dissipate, without overheating
or exploding. (The 'W' stands for 'Watts'.)  It doesn't tell us anything about
their individual resistances. So we don't have enough information to calculate
their series equivalent.
5 0
4 years ago
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w
Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

\theta= (0.20 rad/s)(75 s)=15 rad

In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

v=(0.20 rad/s)(13.5 m)=2.7 m/s

6 0
3 years ago
Why is gravitational force and gravity not the same thing?
vekshin1

Answer:

gravitational force is a force while gravity is a fundamental quantity

Explanation:

3 0
3 years ago
Read 2 more answers
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

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3 years ago
One description of the potential energy of a diatomic molecule is given by the Lennard-Jones potential,U =A/r¹² - B/r⁶ where A a
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