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exis [7]
2 years ago
7

A particle with charge − 3.74 × 10 − 6 C is released at rest in a region of constant, uniform electric field. Assume that gravit

ational effects are negligible. At a time 8.36 s after it is released, the particle has a kinetic energy of 3.17 × 10 − 8 J. At what time t after the particle is released has it traveled through a potential difference of 0.401 V?
Physics
1 answer:
dalvyx [7]2 years ago
4 0

Answer:

57.5022228905 s

Explanation:

V_i = Initial voltage

V_f = Final voltage = 0.401 V

t_1 = Initial time = 8.36 s

t_2 = Final time

K = Kinetic Energy = 3.18\times 10^{-8}\ J

C = Charge = 3.74\times 10^{-6}\ C

Voltage is given by

V=\dfrac{K}{C}\\\Rightarrow V=\dfrac{3.17\times 10^{-8}}{3.74\times 10^{-6}}\\\Rightarrow V=0.00847593582888\ V

We have the relation

\dfrac{V_i}{V_f}=(\dfrac{t_i}{t_f})^2\\\Rightarrow t_f=\sqrt{\dfrac{t_i^2V_f}{V_i}}\\\Rightarrow t_f=\sqrt{\dfrac{8.36^2\times 0.401}{0.00847593582888}}\\\Rightarrow t_f=57.5022228905\ s

The time taken is 57.5022228905 s

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Answer:

Explanation:

Given that

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= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is F_{ma}=BqV---(i)

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Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br}  \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg

Therefore,  the particle's charge-to-mass ratio is 958.1\times10^5C/kg

b)

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Future space rockets might propel themselves by firing laser beams, rather than exhaust gases, out the back. The acceleration wo
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