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3 years ago

Find the speed vfinal of the joined cars after the collision. mastering physics

1 answer:

6 0

<span>Px = 0 Py = 2mV second, Px = mVcosĎ† Py = â€“mVsinĎ† add the components Rx = mVcosĎ† Ry = 2mV â€“ mVsinĎ† Magnitude of R = âš(RxÂ˛ + RyÂ˛) = âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) and speed is R/3m = (1/3m)âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) simplifying Vf = (1/3m)âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) Vf = (1/3)âš((VcosĎ†)Â˛ + (2V â€“ VsinĎ†)Â˛) Vf = (V/3)âš((cosĎ†)Â˛ + (2 â€“ sinĎ†)Â˛) Vf = (V/3)âš((cosÂ˛Ď†) + (4 â€“ 2sinĎ† + sinÂ˛Ď†)) Vf = (V/3)âš(cosÂ˛Ď†) + (4 â€“ 2sinĎ† + sinÂ˛Ď†)) using the identity sinÂ˛(Ď)+cosÂ˛(Ď) = 1 Vf = (V/3)âš1 + 4 â€“ 2sinĎ†) Vf = (V/3)âš(5 â€“ 2sinĎ†)</span>

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Answer:

Explanation:

The path length difference = extra distance traveled

The destructive interference condition is:

$\Delta d = (m+1/2)\lambda$

where m =0,1, 2,3........

So, ←

$\Delta d = (m+1/2)\lamb da9/tex]so [tex]\Delta d = \frac{\lambda}{2}$

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3 years ago

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