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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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For each of the following decays or reactions, name at least. one conservation law that prevents it from occurring.(c) p → π⁺ +
Amanda [17]

(c)p→π⁺₊π⁺₊π

Baryon number is +1 on the left side of the equation, 0 on the

right side. Baryon number is not conserved.

<h3>How do you determine whether a baryon number is conserved?</h3>

  • According to the law of conservation of baryon number, the sum of the baryon numbers of all incoming particles equals the sum of the baryon numbers of all particles produced by the reaction. Energy, and so on, are conserved even if the incoming proton has sufficient energy and charge.
<h3>What is Baryon Number</h3>
  • In particle physics, the baryon number denotes which particles are baryons and which particles are not. Each baryon has a baryon number of 1, and each antibaryon has a baryon number of -1. Other non-baryonic particles have a baryon number of 0. Since there are exotic hadrons like pentaquarks and tetraquarks, there is a general definition of baryon number as:
  • B=1/3(n_{q} -n_q^{-} })
  • wheren_{q} represents the number of quarks and nq represents the number of antiquarks.

To learn more about Baryon Number refer to

brainly.com/question/10358797

#SPJ4

4 0
2 years ago
Which two options are forms of potential energy
blsea [12.9K]

Answer:

e and b

Explanation:

6 0
3 years ago
Read 2 more answers
HELP ME PLEASE !!
kakasveta [241]

Answer:

B 20 sec

Explanation:

if you look at 1992 you can see that it is on 20 sec

4 0
3 years ago
A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately
Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car V_{c} = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

now distance between polive car and speeding car

Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0
3 years ago
5 points
lesantik [10]

Answer:

True

Explanation:

Because everything usually has chemicals

Plz say it's false if i gave out the wrong answer

7 0
3 years ago
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