The De Broglie's wavelength of a particle is given by:
![\lambda=\frac{h}{p}](https://tex.z-dn.net/?f=%20%5Clambda%3D%5Cfrac%7Bh%7D%7Bp%7D%20%20)
where
is the Planck constant
p is the momentum of the particle
In this problem, the momentum of the electron is equal to the product between its mass and its speed:
![p=m_e v=(9.1 \cdot 10^{-31} kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^{-23} kg m/s](https://tex.z-dn.net/?f=%20p%3Dm_e%20v%3D%289.1%20%5Ccdot%2010%5E%7B-31%7D%20kg%29%287.0%20%5Ccdot%2010%5E7%20m%2Fs%29%3D6.4%20%5Ccdot%2010%5E%7B-23%7D%20kg%20m%2Fs%20)
and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:
![\lambda=\frac{h}{p}=\frac{6.6 \cdot 10^{-34} Js}{6.4 \cdot 10^{-23} kg m/s}=1.0 \cdot 10^{-11} m](https://tex.z-dn.net/?f=%20%5Clambda%3D%5Cfrac%7Bh%7D%7Bp%7D%3D%5Cfrac%7B6.6%20%5Ccdot%2010%5E%7B-34%7D%20Js%7D%7B6.4%20%5Ccdot%2010%5E%7B-23%7D%20kg%20m%2Fs%7D%3D1.0%20%5Ccdot%2010%5E%7B-11%7D%20m%20%20%20)
So, the answer is True.
For the answer to the question above, so at the instant, the acceleration of the airplane is southward, the direction of the velocity is also southward. The direction should be the same because it is both a vector quantity and it does not make sense if the direction and acceleration have different direction.
The trains take <u>57.4 s</u> to pass each other.
Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.
therefore,
![v_A=90 km/h\\ v_B=-80 km/h](https://tex.z-dn.net/?f=v_A%3D90%20km%2Fh%5C%5C%20v_B%3D-80%20km%2Fh)
The relative velocity of the train A with respect to B is given by,
![v_A_B=v_A-v_B\\ =(90km/h)-(-80km/h)\\ =170km/h](https://tex.z-dn.net/?f=v_A_B%3Dv_A-v_B%5C%5C%20%3D%2890km%2Fh%29-%28-80km%2Fh%29%5C%5C%20%3D170km%2Fh)
If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.
The trains are a distance d = 2.71 km apart.
Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,
![t= \frac{d}{v_A_B}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7Bd%7D%7Bv_A_B%7D)
Substitute 2.71 km for d and 170 km/h for ![v_A_B](https://tex.z-dn.net/?f=v_A_B)
![t= \frac{d}{v_A_B}\\ =\frac{2.71 km}{170 km/h} \\ =0.01594 h](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7Bd%7D%7Bv_A_B%7D%5C%5C%20%3D%5Cfrac%7B2.71%20km%7D%7B170%20km%2Fh%7D%20%5C%5C%20%3D0.01594%20h)
Express the time in seconds.
![t=(0.01594h)(3600s/h)=57.39s](https://tex.z-dn.net/?f=t%3D%280.01594h%29%283600s%2Fh%29%3D57.39s)
Thus, the trains cross each other in <u>57.4 s</u>.
5.16 meters!!!!!!!!!!!!!!!!!!!!
I think the statement is false. Racewalking involves less impact than running. It <span> is a long-distance discipline within the sport of athletics. Although it is a foot race, it is different from running in that one foot must appear to be in contact with the ground.</span>