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avanturin [10]
3 years ago
14

Classify the waves as longitudinal or transverse

Physics
1 answer:
azamat3 years ago
7 0
Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves. ... A sound wave traveling through air is a classic example of a longitudinal wave.
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Use a yellow (Red + Green) spotlight to make it look black.

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3 years ago
A meter stick is at rest on frictionless surface. A hockey puck is going towards the 30cm mark on the stick and is traveling per
faust18 [17]

Answer:

A)0.306k

B)0.1325k

C)v_{s_{f}} =7.65(-i)+17.35(-j)

D)w=41.64 rads^{-1}

Explanation:

Given:

  • hockey puck is moving towards 30cm mark perpendicular to the stick
  • m_{s} = 0.05kg
  • m_{h} =0.17kg
  • v_{h_{i} } = 9 ms^{-1}
  • after collision the puck is deflected 30°
  • v_{h_{f} } =4.5 ms^{-1}

To find the initial angular momentum about origin which is the 50th mark of the metre scale (It's COM) :                                                                                                                          angular momentum L=mvxr                                      where,                                                                                                                       v  - is the velocity of the puck perpendicular to the radial vector

r  -  is the radius vector

∴

A) L_{i} =  m_{h}r*v_{h_{i} }\\L_{i}= 0.17*\frac{50-30}{100} (-i)*9(-j)\\L_{i} = 0.306 k

B) after collision , it moves 30° from original path;

   and it's speed = \frac{9}{2} =4.5ms^{-1};

   ∴the perpendicular velocity v_{per} = 4.5cos30 = 2.25\sqrt{3}ms^{-1}

⇒L=m_{h}r*v=0.17*0.2(-i)*2.25\sqrt{3}(-j) \\L= 0.1325 k

C) <em><u>since the net external force on the system is zero , the total momentum of the system can be conserved .</u></em>

thus ,

m_{s}v_{s_{i}}+m_{h}v_{h_{i}}=m_{s}v_{s_{f}}+m_hv_{h_{f}}\\0+0.17*9(-j)=0.05*v_{s_{f}}+0.17*(2.25(i)+2.25\sqrt{3}(-j))\\

solving this we get,

⇒v_{s_{f}} =7.65(-i)+17.35(-j)

D) <u><em>since there is no external torque about the system ,the angular momentum can be conserved.</em></u>

L_{h_{i}}= L_{h_{f}} + Iw

where ,

w is the angular velocity of the stick.

I is the moment of inertia of the stick about COM :

I =\frac{m_{s}l^{2}}{12} \\m_{s}=0.05kg\\l=1m\\I = 0.004167 kgm^{2}

∴

⇒0.306k=0.1325k+0.004167w\\w=41.64 rads^{-1}

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