Brian made this table to organize his notes on changes of state. Energy Changa Change of Seate melting freezing deposition condensation gain lose gain Tose Which change of state has the wrong energy change listed?
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Ответ:
Brian XD
Explanation:
Ответ:
The answer is deposition
Explanation: I just did the test also I DO NOT SUPPORT ABBORTION
Ответ:
= 2.52 x 10^ 6 m/s
Explanation:
The force that acts on charged particles between capacitor plates =
F = (q) (Δv) ÷ d
Here, d = distance between the two plates
q = charge of the charged particle
Δv = voltage
Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration
Poting this equation with the first one, we have:
m X a = (q) (Δv) ÷ d
So, the acceleration of a proton when moving towards a negatively charged plate is
a = (q) (Δv) ÷ (d) (m) {proton}
Likewise, the acceleration of an electron when moving towards a positively charged plate is
a = (q) (Δv) ÷ (d) (m) {electron}
Dividing the proton acceleration formula by the electron acceleration formula we have:
a (proton) / a (electron) = m (proton) / m(electron)
inserting equation of motion to get distance, s
s = ut + 1/2 at^2
recall that electron travel distance, d/2
d/2 = 1/2 at^2
making t the subject of the formula
we have, t =√(d ÷ a(electron))
The distance of proton:
d/2 = ut + 1/2 at^2 [proton}
put d/2 = ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))
Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))
since acceleration wasn't given in the question, lets use mass(elect
ron) ÷ mass(proton) rather than use (a(proton) + a(electron))
Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))
Note, e = 1.60 x 10^-19
m(electron) = 9.11 X 10^-31
m(proton) = 1.67 X 10^-27
Input these values into the formula above, initial speed, UI =
= 2.52 x 10^ 6 m/s