5

4 reckecf ;s flyi5 #olgh space tal ls 4 ckech < 44 "Msnehe frete" 6 < 560c,J4) Ey 1,000,p00 VmD >vBz = 0, Od5t * Wha+ Ts +z magnxtud of te magnct...

Question

4 reckecf ;s flyi5 #olgh space tal ls 4 ckech < 44 "Msnehe frete" 6 < 560c,J4) Ey 1,000,p00 VmD >vBz = 0, Od5t * Wha+ Ts +z magnxtud of te magnctic ficld as MeaSured 6} c Sume observer ? B= T

4 reckecf ;s flyi5 #olgh space tal ls 4 ckech < 44 "Msnehe frete" 6 < 560c,J4) Ey 1,000,p00 Vm D >v Bz = 0, Od5t * Wha+ Ts +z magnxtud of te magnctic ficld as MeaSured 6} c Sume observer ? B= T



Answers

Go As in Fig. $37-9,$ reference frame
$S^{\prime}$ passes reference frame $S$
with a certain velocity. Events 1 and 2 are to
have a certain temporal scparation $\Delta
t^{\prime}$ according to the $S^{\prime}$
observer. However, their spatial scparation
$\Delta x^{\prime}$ according to that
observer has not been set yet. Figure $37-
24$ gives their temporal separation $\Delta
t$ according to the $S$ observer as a
function of $\Delta x^{\prime}$ for a range of
$\Delta x^{\prime}$ values. The vertical axis
scale is sct by $\Delta t_{a}=6.00 \mu
\mathrm{s}$. What is $\Delta t^{n}$

Problem Number 20 Tractor Real activity from a table. Then we can just write equation. Delta T will be equal to lost e times Gamma factor times Dull tax dish or C square plus Gama Delta T Dish the coefficient off a delta X Here, the coefficient of the Delta X Here is the slope from the graph we can see. So it's slope, which is which is 4.0, biker seconds microseconds. So this is new micro seconds or 40 meter from here. Then we can find a better. So the better we will get here is we or C, which is zero point 94 line. Then the gamma will have is 3.16 So the gamma we obtained from over relation offer electricity from observation. Then we can right Delta D dish as delta T or gamma gamma by substituting these values. Then we get Delta tease 64 and three times 10 to the power minus seven seconds. End off the problem. Thank you.

In this problem. On the topic of relativity, we are told that a meter stick is moving at 0.9 C on 90% the speed of light relative to the earth's surface. Edin approaches an observer that is addressed with respect to the art surface, and we want to know the meat sticks length as measured by this observer. And then we want to know how the answer to part they would change if the Observer started running toward the meter stick. Now we first need to calculate the length of the meter meter stick measured by the Observer, moving at 0.90 to the meter stick. So this land is L and L is the rest land L P divided by gamma using Lawrence transformations. So this is LP times the square root of one minus the oversee old squared. So we know this is arrest length of 1 m multiplied by the square root of one minus zero 0.9 squared, so calculating we get the length of the meter stick with respect to the Observer to be zero 0.436 m. So that's the observed length of the meter stick. That's moving at 0.9 times the speed of light. Now, if the observer moves relative to the earth in the direction opposite to the emotion of the meter stick and the velocity of the observer relative to the media stick is greater than that in part A. So the measured length of the meter stick will then become less than 0.436 meters, and under these conditions it will be so small that it would be essentially an observable.

Answer for A. Who is at rest in the laboratory is studying a particle that is moving through the laboratory at a speed of 0.6 to foresee and determines its lifetime to be 159 nanoseconds. Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory? Observer. For part B. Observer B. Who is traveling parallel to the particle at a speed of 0.624 C. Observes the particle at rest and measures its lifetime to be 100 and 24 nanoseconds According to be. How far apart are the two markers in the laboratory? So for part A we have a delta X. A. Is equal to V. Delta T. A. We have adult IT T. A. Is equal to 159 nanoseconds. So this gives us 29.7 m. And for part B we have delta X. B. Is equal to be delta T. B. And they told us still to TB is equal to 124 nanoseconds. So this gives us 23.2 m.

Answer for A. Who is at rest in the laboratory is studying a particle that is moving through the laboratory at a speed of 0.6 to foresee and determines its lifetime to be 159 nanoseconds. Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory? Observer. For part B. Observer B. Who is traveling parallel to the particle at a speed of 0.624 C. Observes the particle at rest and measures its lifetime to be 100 and 24 nanoseconds According to be. How far apart are the two markers in the laboratory? So for part A we have a delta X. A. Is equal to V. Delta T. A. We have adult IT T. A. Is equal to 159 nanoseconds. So this gives us 29.7 m. And for part B we have delta X. B. Is equal to be delta T. B. And they told us still to TB is equal to 124 nanoseconds. So this gives us 23.2 m.


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