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Firlakuza [10]
3 years ago
13

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

ative plate with a speed of 45000 m/s.
What will be the final speed of an electron released from rest at the negative plate?
Express your answer to two significant figures and include the appropriate units
V=_________
Physics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:

2.1x10^6m/s

Explanation:

One electron has a charge of –1.602e-19 C

mass of electron is 9.1e-31 kg

mass of proton is 1.6726e−27 kg

mass ratio is 1.6726e−27 / 9.1e-31 = 1838

The force is constant, F

distance is constant, d

a = F/m

a increases by a factor 1838, as m decreases by that factor

a = a₀1838

v₀² = 2a₀d

v² = 2a₀d1838

v²/v₀² = 2a₀d1838 / 2a₀d = 1838

v² = 1838v₀² = 1838(45000)²

v = 45000√1838 = 2.1e6 m/s

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Answer:

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A person takes a trip, driving with a constant speed of 94.5 km/h except for a 22.0 min rest stop. If the person's average speed
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From the average speed you can fix an equation:

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The distance is the same in both equations, then you have:

[t +22/60] * 65.1 = 94.5 t

Now you can solve for t.

65.1t + 22*65.1/60 = 94.5t

94.5t - 65.1t = 22*65.1/60

29.4t = 23.87

t = 23.87 / 29.4

t  = 0.812 hours

distance = 94.5 km/h * 0.812 h = 76.7 km

Answers: 1) 0.81 hours, 2) 76.7 km


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Explanation:

this is my answer this is helpful for you

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