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What is the speed of a proton that has been accelerated from rest through a potential difference of -900

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Answer:

To answer this question, we need to know the mass of the proton and the value of the potential difference it has been accelerated through.

The mass of a proton is approximately 1.67 x 10^-27 kg.

The potential difference is given as -900 volts. Since the potential difference is negative, we can assume that the proton is accelerating towards a negatively charged plate.

We can use the equation for the kinetic energy of an object to find the speed of the proton:

KE = (1/2)mv^2

where KE is the kinetic energy, m is the mass of the proton, and v is its speed.

We also know that the potential energy gained by the proton is equal to the kinetic energy it possesses after being accelerated. This potential energy can be calculated as:

PE = qV

where q is the charge of the proton and V is the potential difference.

Since the charge of a proton is 1.6 x 10^-19 coulombs, we get:

PE = (1.6 x 10^-19 coulombs)(-900 volts)

PE = -1.44 x 10^-16 joules

Setting KE = -PE, we get:

(1/2)mv^2 = -1.44 x 10^-16 joules

v^2 = -2(1.44 x 10^-16 joules)/(1.67 x 10^-27 kg)

v = 3.36 x 10^7 m/s

So the speed of the proton after being accelerated through a potential difference of -900 volts is approximately 3.36 x 10^7 meters per second.

Step-by-step explanation:

User GGalizzi
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