Explanation:
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During hockey practice, two pucks are sliding across the ice in the same direction. At one instant, a 0.18-kg puck is moving at
1 answer:
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Answer:
24.348mm
Explanation:
NB: I'll be attaching pictures so as to depict missing mathematical expressions or special characters which are not easily found on keyboards
K = d / €^n
Note : d represents the greek alphabet epsilion.
K = 345 / 0.02⁰.²² = 816mPa
The true strain based upon the stress of 414mPa =
€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576
However the true relationship between true strain and length is given by
€ = ln(Li/Lo)
Making Li the subject of formula by rearranging,
Li = Lo.e^€
Li = 520e⁰.⁰⁴⁵⁷⁶
Li = 544.348mm
The amount of elongation can be calculated from
Change in L = Li - Lo = 544.348 - 520 change in L = 24.348mm.
Answer:
The final size is approximately equal to the initial size due to a very small relative increase of in its size
Solution:
As per the question:
The energy of the proton beam, E = 250 GeV =
Distance covered by photon, d = 1 km = 1000 m
Mass of proton,
The initial size of the wave packet,
Now,
This is relativistic in nature
The rest mass energy associated with the proton is given by:
This energy of proton is
Thus the speed of the proton, v
Now, the time taken to cover 1 km = 1000 m of the distance:
T =
T =
Now, in accordance to the dispersion factor;
Thus the increase in wave packet's width is relatively quite small.
Hence, we can say that:
where
= final width