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Alex_Xolod [135]

3 years ago

7

An electron of mass 9.11×10-31kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to t

he accelerating grid, which is a distance 1.30cm away. It reaches the grid with a speed of 2.50×106m/s . The accelerating force is constant.
a)Find the acceleration.
b)Find the time to reach the grid.
c)Find the net force. (You can ignore the gravitational force on the electron).

1 answer:

SOVA2 [1]3 years ago

7 0

Answer:

(A) Acceleration will be $240.3846\times 10^{12}m/sec^2$

(b) Time taken will be $1.4\times 10^{-8}sec$

(c) Force will be $2189.9\times 10^{-19}N$

Explanation:

We have given that electron starts from rest so initial velocity u = 0 m/sec

Final velocity $v=2.50\times 10^6m/sec$

Mass of electron $m=9.11\times 10^{-31}kg$

Distance traveled by electron $s=1.30cm =0.013m$

From third equation of motion we know that $v^2=u^2+2as$

(a) So $(2.5\times 10^6)^2=0^2+2\times a\times 0.013$

$a=240.3846\times 10^{12}m/sec^2$

(b) From first equation of motion we know that v = u+at

So $2.50\times 10^6=0+240.3846\times 10^{12}t$

$t=0.014\times 10^{-6}=1.4\times 10^{-8}sec$

(c) From newton's law we know that force

$F=ma=9.11\times 10^{-31}\times 240.3846\times 10^{12}=2189.9\times 10^{-19}N$

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How'd I do this?

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$W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

<u>Explanation:</u>

The parametric representation of a line segment joining the points (a,b,c) and (l,m,n) is

r(t) = (1-t) . (a,b,c) + t . (l, m, n) where t ∈ |0, 1|

So, the parametric representation of a line segment joining the points (3,0,0) and (3,2,5) is

r(t) = (1 - t) . (3,0,0) + t . (3,2,5) where t ∈ |0, 1|

r(t) = (3(1 - t), 0, 0) + (3t, 2t, 5t) where t ∈ |0, 1|

r(t) = (3, 2t, 5t)

Given:

$F(x, y, z) = \frac{Kr}{|r|^3} \\\\F(x, y, z) = \frac{K}{(x^2 + y^2 + z^2)^3^/^2} (x, y, z)\\\\F(r(t)) = \frac{K}{(3^2 + (2t)^2 + (5t)^2)^3^/^2} (3, 2t, 5t)\\\\F(r(t)) = \frac{K}{(9 + 29t^2)^3^/^2} (3, 2t, 5t)$

dr = (0, 2, 5) dt

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$W = \int\limits^1_0 {\frac{K}{(9 + 29t^2)^3^/^2} } (3, 2t, 5t) . (0, 2, 5)\, dt\\ \\ = \int\limits^1_0 {\frac{K ( 4t + 25t)}{(9 + 29t^2)^3^/^2} } \, dt\\\\\\$

$W = \frac{1}{2}\int\limits^1_0 {\frac{K(29t)}{(9 + 29t^2)^3^/^2} } \, dt \\ \\$

Substitute = 9 + 29t² = u, 92tdt = du

Limit changes from 0→1 to 9 → 38

$W = \frac{K}{2} \int\limits^3_9 {\frac{du}{u^3^/^2} } \,\\\\$

On solving this, we get:

$W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

Therefore, work done is $W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

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