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exis [7]
3 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]3 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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An object with total mass mtotal = 14.6 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.9 k
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3) and 4)

As momentum is a vector, if it is magnitude must be 0, this means that all his components must be 0 too.

So, we can write two equations, one for the x-component, and other for the y-component, as follows:

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6) As initial kinetic energy is 0, as  the mass was at rest, the increase in the kinetic energy is obtained simply adding the kinetic energy of every piece of mass gained after explosion, as follows:

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K= 3.62 KJ

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