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Schach [20]
3 years ago
6

According to quantum physics, measuring velocity of a tiny particle with an electromagnet

Physics
1 answer:
lidiya [134]3 years ago
4 0

Answer:

Option A.

Explanation:

In quantum physics <u>there is a law to relate the position and the momentum of the particle</u>, it says that if we know with precision where is a quantum particle, we can not know the momentum of this particle, in other words, the velocity of the particle. So, when we measure the velocity of the particle we find the correct value of the particle, but we can not determine with accuracy where is the particle. This law is known as the Heisenberg's uncertainty principle and, its expressed as follows:    

\Delta x \Delta p \geq \frac{h}{4 \pi}

<em>where Δx: is the position's uncertainty, Δp: is the momentum's uncertainty and h: is the Planck constant.</em>  

Therefore, the correct answer is A: measuring the velocity of a tiny particle with an electromagnet has no effect on the velocity of the particle. It only affects the determination of the particle's position.      

I hope it helps you!

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A 5000kg freight car moving at 2 m/s East collides with a 10,000kg freight car at rest. Upon collision, they got stuck and moved
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An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
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Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

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• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

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Hope this helps :)
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