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lesya [120]
3 years ago
12

Show that any three linear operators A, B, and Ĉ satisfy the following (Ja- cobi) identity (10 pt) [[A, B] Ĉ] + [[B,C), A] + [[C

, A, B) = 0
Physics
1 answer:
DerKrebs [107]3 years ago
7 0

Answer:

Three linear operators A,B, and C will satisfy the condition [[A, B],C] + [[B,C), A] + [[C, A], B] = 0.

Explanation:

According to the question we have to prove.

[[A, B],C] + [[B,C), A] + [[C, A], B] = 0

Now taking Left hand side of the equation and solve.

[[A, B],C] + [[B,C), A] + [[C, A], B]

Now use commutator property on it as,

=[A,B] C-C[A,B]+[B,C]A-A[B,C]+[C,A]B-B[C,A]\\=(AB-BA)C-C(AB-BA)+(BC-CA)A-A(BC-CB)+(CA-AC)B-B(CA-AC)\\=ABC-BAC-CAB+CBA+BCA-CAB-ABC+ACB+CAB-ACB-BCA+BAC\\=0

Therefore, it is proved that [[A, B],C] + [[B,C), A] + [[C, A], B] = 0.

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What is the impulse needed to stop a 45-kg boy who is running at 6 m/s in 3 seconds?
Tju [1.3M]
Impulse = change of momentum
Impulse = 45 x 6 = 270 Ns
6 0
3 years ago
Which best explains how the diffraction pattern observed in young’s experiment supports the wave theory of light?.
KIM [24]

Answer: the airy pattern can only arise from wave propagation

Explanation:if particles went in straight lines through a slit, they would progate linearly and not interfere. The airy pattern arises from diffraction as waves interfere, producing peaks (constructive interference where peaks of waves from each slit coincide) and troughs (destructive interference where peaks and troughs of waves from each slit cancel out). If intensity rather than field is measured nodes occur where 0 values line up instead of troughs

6 0
2 years ago
Inside a NASA test vehicle, a 3.50-kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The forc
erastova [34]
F(of spring)=230x=ma=3.5(5)=17.5=230x; x=0.07m.
3 0
3 years ago
A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b
jonny [76]

Answer:

Part a)

I = 3.16 \times 10^{-5} W/m^2

Part b)

P_o = 0.162 Pa

Explanation:

Part a)

Level of sound = 75 dB

now we know that

L = 10 Log\frac{I}{I_0}

here we know that

I_0 = 10^{-12} W/m^2

now we have

75 = 10 Log(\frac{I}{10^{-12}})

I = 3.16 \times 10^{-5} W/m^2

Part b)

Intensity of sound wave is given as

I = \frac{1}{2}\rho A^2\omega^2 c

here we know that

A = \frac{P_o}{Bk}

so we have

I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c

I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}

now we know

\rho = 1.2 kg/m^3

c = 340 m/s

B = 1.4 \times 10^5 Pa

now we have

3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}

P_o = 0.162 Pa

5 0
3 years ago
A 0.4000 kg sample of methanol at 16.0ºC is mixed with 0.4000 kg of water at 85.0ºC. Assuming no heat loss to the surroundings,
AVprozaik [17]

Answer:

T_finalmix = 59.5 [°C].

Explanation:

In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water to methanol, at the end the temperature of the water and methanol must be equal once the thermal balance is achieved.

Q_{water}=Q_{methanol}

where:

Q_{water}=m_{water}*Cp_{water}*(T_{waterinitial}-T_{final})

mwater = mass of the water = 0.4 [kg]

Cp_water = specific heat of the water = 4180 [J/kg*°C]

T_waterinitial = initial temperature of the water = 85 [°C]

T_finalmix = final temperature of the mix [°C]

Q_{methanol}=m_{methanol}*Cp_{methanol}*(T_{final}-T_{initialmethanol})

Now replacing:

0.4*4180*(85-T_{final})=0.4*2450*(T_{final}-16)\\142120-1672*T_{final}=980*T_{final}-15680\\157800=2652*T_{final}\\T_{final}=59.5[C]

4 0
3 years ago
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