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ziro4ka [17]
4 years ago
13

Muons are elementary particles that are formed high in the atmosphere by the interactions of cosmic rays with atomic nuclei. Muo

ns are radioactive and have average lifetimes of about two-millionths of a second. Even though they travel at almost the speed of light, they have so far to travel through the atmosphere that very few should be detected at sea level - at least according to classical physics. Laboratory measurements, however, show that muons in great number do reach the earth's surface. What is the explanation?
Physics
1 answer:
uysha [10]4 years ago
5 0

Answer:

<em>Muons reach the earth in great amount due to the relativistic time dilation from an earthly frame of reference.</em>

Explanation:

Muons travel at exceedingly high speed; close to the speed of light. At this speed, relativistic effect starts to take effect. The effect of this is that, when viewed from an earthly reference frame, their short half life of about two-millionth of a second is dilated. The dilated time, due to relativistic effects on time for travelling at speed close to the speed of light, gives the muons an extended relative travel time before their complete decay. So <em>in reality, the muon do not have enough half-life to survive the distance from their point of production high up in the atmosphere to sea level, but relativistic effect due to their near-light speed, dilates their half-life; enough for them to be found in sufficient amount at sea level.  </em>

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satela [25.4K]

Answer:

The  horizontal displacement is  Adj  =  114.71 \ m

Explanation:

From the question we are told that  

    The  angle at which the string is maintained is  \theta  =  41 ^o

      The length of string reeled in is  l  = 152 \ m

     

Using the SOHCAHTOA formula

   We have that the hypotenuse(Hyp) is  l  =  152

Hence the  horizontal displacement of the kite  which is the Adjacent(Adj)  can be evaluated as  

     cos \theta  =  \frac{Adj}{Hyp }

substituting values

     cos(41)  =  \frac{Adj}{152}

=>   Adj  =  114.71 \ m

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Levart [38]
Answer:
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Calculate the pressure when a force of 200N acts on an area of 0.05cm2<br> .
deff fn [24]

FORCE ACTING ON THE BODY , F = 200N

AREA ON WHICH THE FORCE IS ACTING , A = 0.05 CM²

= 0.05*(10⁻²)

= 0.05*10⁻² M²

TO CONVERT CM TO M * BY 10⁻²

PRESSURE = FORCE / AREA

PRESSURE = 200/0.05*10⁻⁴

= 200*10⁴/0.05

= 200*10⁴*10²/5

= 20*10⁵/5

= 4*10⁵ N/M².

HOPE THIS HELPS.

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