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vladimir2022 [97]
3 years ago
12

If a person traveled from Sacramento to Los Angeles by car, their distance would be 384.9 miles. What would their displacement b

e? _____________________ If that same person returned to Sacramento from Los Angeles by the same route in the same car, what would the total round trip distance be? 400 miles What would their total displacement be? _____________________
Physics
1 answer:
kumpel [21]3 years ago
3 0

Answer:

Displacement from Sacramento to Los Angeles is 384.9 miles

Total path length is given as 769.8 miles

Displacement of complete round trip is ZERO

Explanation:

As we know that the displacement is the change in the position from initial to final in a straight line

Here we know that the car travel from Sacramento to Los Angeles

So the straight line distance between two is the displacement

so we have

displacement = 384.9 miles

Now we know that he returns from Los Angeles to Sacramento by same path

So total distance moved by it is same as total path length

d = 384.9 + 384.9

d = 769.8 miles

Also we know that initial and final positions are same so displacement is ZERO for complete round trip.

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The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel
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Answer:

= 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          

7 0
3 years ago
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