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Paul [167]
3 years ago
14

Classical mechanics is an extremely well tested model. Hundreds of years worth of experiments, as well as most feats of engineer

ing, have verified its validity. If special relativity gave very different predictions than classical physics in everyday situations, it would be directly contradicted by this mountain of evidence. In this problem, you will see how some of the usual laws of classical mechanics can be obtained from special relativity by simply assuming that the speeds involved are small compared to the speed of light.Two of the most surprising results of special relativity are time dilation and length contraction, namely, that measured intervals in time and space are not absolute quantities but instead appear differently to different observers. The equations for time dilation and length contraction can be written t=?t0 and l=l0/?, where?=11?u2c2?.Part AFind the first two terms of the binomial expansion for ?.Express your answer in terms of u and c.Hints? = 1+12(uc)2 … SubmitMy AnswersGive UpCorrectYou can see that ??1 if u?c, as is the case in most situations. If you set ?=1 in the equations for time dilation and length contraction you recover the equations of classical physics, which state essentially that there is no time dilation or length contraction. Therefore, we don't see any appreciable length contraction or time dilation in everyday life.Part BConsider a case involving a speed that is fast compared to those encountered in our everyday life: a spy plane moving at 1500m/s. Find the deviation from classical physics (??1) that relativity predicts at this speed. Use only the first two terms of the binomial expansion, as your calculator may not be able to handle the necessary number of digits otherwise.Express your answer to four significant figures.??1 = 1.250×10?11SubmitMy AnswersGive UpCorrectIf you lived for 70 years in such a spy plane moving at 1500m/s, this would amount to about 28ms of cumulative time difference between you and people who lived at rest relative to the earth when you finally landed. Thus, it is not surprising that relativistic effects are not observed in everyday life, or even at the fringes of everyday life. By using atomic clocks, which can measure time accurately to one part in 1013 or better, the time dilation at the normal speed for an airliner has been verified.Part CNow, consider the relativistic velocity addition formula:speed=v+u1+vuc2.If v=u=0.01c=1% of c, what is the relativistic sum of the two speeds?Express your answer as a percentage of the speed of light to five significant figures.

Physics
1 answer:
djyliett [7]3 years ago
7 0

Answer:

The Answer is 0.019998c

Explanation:

Please see the attached Picture for the answer.

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Stephen is worried he has a fever. His dad is a scientist though and only has thermometer that measures in kelvin and is measure
MaRussiya [10]

Answer: im trying to find the same answer too

Explanation

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The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of
LekaFEV [45]

Answer:

   (h₁-h₂) = 2.30 10² m

Explanation:

The pressure depends on the height with the formula

          P = P_atm + rho g h

Let's apply this expression for the building

         P₁ = P_atm + rho_air g h₁

        P₂ = P_atm + rho_air g h₂

Subtract

        P₁ - P₂ = roh_air g (h₁ –h₂)

         

         

The measured pressure is in mm Hg to take this unit to units of pressure must be multiplied by the density of mercury and the acceleration of gravity

           P₁- P₂ = rho_Hg g (h₁-h₂) _Hg

         rho_Hg g (h₁-h₂) _Hg = roh_air g (h₁ –h₂)

          (h₁ –h₂) = rho_Hg / rho_air (h₁-h₂) _ Hg

Let's calculate

         (h₁-h₂) = 13600 / 1.18 (695-675)

         (h₁-h₂) = 2.30 10⁵ mm

Let's reduce to meter

         (h₁-h₂) = 2.30 10⁵ mm (1 m / 10³ mm)

         (h₁-h₂) = 2.30 10² m

4 0
3 years ago
Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the pos
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Answer:

E = (0, 0.299) N

Explanation:

Given,

  • Charge q_1\ =\ 8.0\ nC
  • Charge q_2\ =\ 6.0\ nC
  • Distance of the first charge from the origin = (16m, 0)
  • Distance of the second charge from the origin = (-9, 0)
  • Point where the electric field required = (0, 12m)

Let \theta_1\ and\ theta_2 be the angle of the electric fields by first and second charge at the point A.

\therefore sin\theta_1\ =\ \dfrac{12}{20}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{20}\ \right )\\\Rightarrow \theta_1\ =\ 36.87^o\\\\\therefore sin\theta_1\ =\ \dfrac{12}{9}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{9}\ \right )\\\Rightarrow \theta_1\ =\ 53.13^o\\

Electric field by charge q_1 at point A,

F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 8\times 10^{-9}}{20^2}\\\Rightarrow F_1\ =\ 0.18\ N/C

Electric field by the charge q_2 at point A,

F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 6.0\times 10^{-9}}{16^2}\\\Rightarrow F_1\ =\ 0.24\ N/C

Now,

Net electric field in horizontal direction at point AF_x\ =\ F_{1x}\ +\ F_{2x}\\\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\\\Rightarrow F_x\ =\ 0.18\times( -cos36.87^o)\ +\ 0.24\times cos53.13^o\\\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\\\Rightarrow F_x\ =\ 0\ N/C

Net electric field in vertical direction at point A.

F_y\ =\ F_{1y}\ +\ F_{2y}\\\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\\\Rightarrow F_y\ =\ 0.18\times sin36.87^o\ +\ 0.24\times sin53.13^o\\\Rightarrow F_y\ =\ 0.180\ +\ 0.192\\\Rightarrow F_y\ =\ 0.299\ N/C

Hence, the net electric field  at point A,

F\ =\ ( 0, 0.299 )\ N/C.

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Answer:

my favourite song is hym to the sea. That song reminds me of the innocent people who lost their lives in titanic a century ago. If I was the writer of that song ,i would not change the title of that song because I can feel the sound of ocean when I listen to that song.

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