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nikitadnepr [17]
3 years ago
6

What is the amplitude of the resulting wave when the two waves meet at point X?

Physics
1 answer:
IrinaK [193]3 years ago
8 0

Answer:

Resulting amplitude, A = 0 cm

Explanation:

The phenomenon of interference occurs when two travel and meet at the same point. There are two conditions for interference. The first one is constructive and the other one is destructive.

In this case, two waves have same amplitude but they are not in same phase such that they meet at point X and destructive interference occurs.

Let A is the amplitude of the resulting wave such that,

A=A_1-A_2

A=2\ cm-2\ cm

A = 0

So, the amplitude of the resulting wave is 0 cm. Hence, the correct option is (A).

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Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
Oksana_A [137]

Answer:

av=0.333m/s, U=3.3466J

b.

v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s

Explanation:

a. let m_A be the mass of block A, andm_B=10.0kg be the mass of block B. The initial velocity of A,\rightarrow v_A_1=2.0m/s

-The initial momentum =Final momentum since there's no external net forces.

pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

v_A_1-v_B_1=v_{B2}-v_{A2}

-Applying the conservation of momentum. The blocks have the same velocity after collision:

v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s

#Total Mechanical energy before and after the elastic collision is equal:

K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0

We plug these values in the equation:

m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s

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3 years ago
At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise betw
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To develop this problem it is necessary to apply the concepts related to the Dopler effect.

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