In which case is the speed of laser light the slowest relative to the observer? a person walking in the street and who sees lase
1 answer:
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Answer:
<u>For proton:</u>
A. The proton is released from Vb (highest potential)
B. v = 2.9x10⁴ m/s
<u>For electron:</u>
A. The electron is released from Va (lowest potential)
B. v = 1.2x10⁶ m/s
Explanation:
<u>For a proton we have</u>:
A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.
Having that:
Va = 1.51 V and Vb = 5.81 V
We can see that the proton moves from Vb to Va, hence the proton was released from Vb.
B. We now that the work done by an electric field is given by:
(1)
Where:
q: is the proton's charge = 1.6x10⁻¹⁹ C
V: is the potential
Also, the work is equal to:
(2)
Where:
K: is the kinetic energy
m: is the proton's mass = 1.67x10⁻²⁷ kg
: is the velocity in the point a
: is the velocity in the point b = 0 (starts from rest)
Matching equation (1) with (2) we have:
<u>For an electron we have</u>:
A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.
B. The speed is:
Since = 0 (starts from rest) and
= 9.1x10⁻³¹ kg (electron's mass), we have:
I hope it helps you!