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ycow [4]
3 years ago
5

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

tron be moving when it is 3.00 cm from the proton?
Physics
1 answer:
lana66690 [7]3 years ago
8 0

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

  • x = initial distance of the electron from the proton = 6 cm = 0.06 m
  • y = initial distance of the electron from the proton = 3 cm = 0.03 m
  • u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

  • m = mass of an electron = 9.1\times 10^{-31}\ kg
  • v = final velocity of the electron
  • e = magnitude of charge on an electron = 1.6\times 10^{-19}\ C
  • p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

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Water flows over a section of Niagara Falls at the rate of 1.1 × 106 kg/s and falls 50.0 m. How much power is generated by the f
Kryger [21]
<h3>Answer:</h3>

5.395 × 10^8 Watts

<h3>Explanation:</h3>

<u>We are given;</u>

  • Rate of flow is 1.1 × 10^6 kg/s
  • Distance is 50.0 m
  • Gravitational acceleration is 9.8 m/s²

We are required to calculate the power that is generated by the falling water

  • Power is the rate of work done
  • It is given by dividing the energy or work done by time
  • Power = Work done ÷ time

But; work done = Force × distance

Therefore;

Power = (F × d) ÷ time

The rate is 1.1 × 10^ 6 Kg/s

But, 1 kg = 9.81 N

Therefore, the rate is equivalent to 1.079 × 10^7 N/s

Thus,

Power = Rate (N/s) × distance

           = 1.079 × 10^7 N/s × 50.0 m

           = 5.395 × 10^8 Watts

The power generated from the falling water is 5.395 × 10^8 Watts

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Question 7
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Power = energy/time=20/4=5.0
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A +13.4 nC charge is located at (0,9.4) cm and a -4.23 nC charge is located (4.99, 0) cm. Where would a -14.23 nC charge need to
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In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge

E1 = (9x10^9)(13.4x10^-9)/(9.4x10^-2)^2 = 13648.7 V/m towards the negative y-axis

E2 = (9x10^9)(4.23x10^-9)/(4.99x10^-2)^2 = 15289.1 V/m towards the positive x-axis

The red arrow shows the direction of which the electric field points.

To make the electric field at the origin 0, we must find a location where q3 = the magnitude of q1 and q2

Etotal = sqrt(E1+E2) = 20494.97 V/m

E3 = 20494.97 = (9x10^9)(14.23x10^-9)/(d)^2

d = 0.079 m = 7.9 cm

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