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evablogger [386]
3 years ago
7

A proton is traveling horizontally to the right at 4.60×106 m/s . Part A Find (a)the magnitude and (b) direction of the weakest

electric field that can bring the proton uniformly to rest over a distance of 3.50 cm . Part C How much time does it take the proton to stop after entering the field? Part D What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?
Physics
1 answer:
Feliz [49]3 years ago
3 0

Answer:

Explanation:

u = 4.6 x 10^6 m/s

Let E be the electric field

s = 3.5 cm = 0.035 m

v = 0

a = qE / m

So use third equation of motion

v^2 = u^2 - 2 a s

0 = (4.6 x 10^6)^2  - 2 x q E / m x 0.035

21.16 x 10^12 = 2 x 1.6 x 10^-19 x E / (1.67 x 10^-27 x 0.035)

E = 3865 N/C

(a) The magnitude of electric field is 3865 N/C

(b) the direction of electric field is opposite to the direction of motion of proton, i.e., towards left.

(c) Let t be the time taken

v = u + a t

0 = 4.6 x 10^6 - (1.6 x 10^-19 x 3865) t / (1.67 x 10^-27)

t  = 1.24 x 10^-5 sec

(d) For electron, the direction of electric field is same the direction of electron, i.e., rightwards.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = (4.6 x 10^6)^2  - 2 x q E / m x 0.035

21.16 x 10^12 = 2 x 1.6 x 10^-19 x E / (9.1 x 10^-31 x 0.035)

E = 2.1 N/C

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Based off the word "conserved" I would say

A. Conservation of Momentum.

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<h2>♨ANSWER♥</h2>

length of V-50 = 49mm

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The minor measurement of the vernier scale is 0.02mm.

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What is the period of a sound wave having a frequency of 340 Hz
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Hope it helps and have a wonderful day!

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As altitude increases in the troposphere and stratosphere, the air temperature does what?
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Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour
Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 20

t₁= 33.4   years

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2 years ago
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