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zavuch27 [327]
3 years ago
10

A proton and an electron are held in place on the x axis. The proton is at x = -d, while the electron is at x = +d. They are rel

eased simultaneously, and the only force that affects their motions is the electrostatic force of attraction that applies to the other. Which particle reaches the origin first? Give your reasoning.
Physics
1 answer:
Over [174]3 years ago
6 0
The protons and electrons are held in place on the x axis.
The proton is at x = -d and the electron is at x = +d. They are released at the same time and the only force that affects movement is the electrostatic force that is applied on both subatomic particles. According to Newton's third law, the force Fpe exerted on protons by the electron is opposite in magnitude and direction to the force Fep exerted on the electron by the proton. That is, Fpe = - Fep. According to Newton's second law, this equation can be written as
                               Mp * ap = -Me * ae
where Mp and Me are the masses, and ap and ae are the accelerations of the proton and the electron, respectively. Since the mass of the electron is much smaller than the mass of the proton, in order for the equation above to hold, the acceleration of the electron at that moment must be considerably larger than the acceleration of the proton at that moment. Since electrons have much greater acceleration than protons, they achieve a faster rate than protons and therefore first reach the origin.
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Answer: Taking into account sound is a wave, we can use the information of the displacement (generally given as a graph) to find the wavelength and frequency, then we can calculate the speed with the formula of the speed of a wave.

Explanation:

If we have the displacement graph of the sound wave, we can find its amplitude, its wavelength and period (which is the inverse of frequency).

Now, if we additionally have the frequency as data, we can use the equation of the speed of a wave:

s=\lambda f

Where:

s is the speed of the sound wave

\lambda is the wavelength

f is the frequency

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Which image illustrates refraction?
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Answer:

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3 years ago
1. Determine the energy released per kilogram of fuel used.
Serga [27]

The formula for energy release per kilogram of fuel burned is energy release per kg=6.702*10-13. and 19. J 1 Mev = 1.602 X 10 T

Calculate the energy in joules per kilogram of reactants given MeV per reaction. Energy is the ability or capacity to perform tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
Think of a mole of plutonium-239 (molar mass: 239 grams) as a mole of "reactions."

Energy used in the US per person annually = 3-5 X 1011
Population (number of people) = 3.108The required mass of the fuel is 3.5x1011 x3-1x10 8x 10)/6.703 X1013 kg. the mass required: 1.62 x 1033 kg Mev in Joules 6 is equal to 101.60*I0-
19. J 1 Mev = 1.602 X 10 T, which translates to 1.602*1013/2.39x10-3 energy release per kilogram, or 6.702*10-13.

To learn more about Energy please visit -
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8 0
2 years ago
Which substance(s) will sink if placed in a fluid that has a density of 1.19 g/ml? (List all)
olga2289 [7]

Answer:

Corn Syrup

Explanation:

7 0
2 years ago
A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha
Airida [17]

Answer:

  • The work made by the gas is 7475.69 joules
  • The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas  its defined as:

dW =  P \ dv

We can solve this by integration:

\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

P \ V = \ n \ R \ T

P = \frac{\ n \ R \ T}{V}

This give us

\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv

As n, R and T are constants

\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv

\Delta W= \ n \ R \ T  \left [ ln (V) \right ]^{v_2}_{v_1}

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ln (\frac{v_2}{v_1})

But the volume is:

V = \frac{\ n \ R \ T}{P}

\Delta W = \ n \ R \ T  ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )

\Delta W = \ n \ R \ T  ln(\frac{P_1}{P_2})

Now, lets use the value from the problem.

The temperature its:

T = 27 \° C = 300.15 \ K

The ideal gas constant:

R = 8.314 \frac{m^3 \ Pa}{K \ mol}

So:

\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K  ln (\frac{20 atm}{1 atm})

\Delta W = 7475.69 joules

<h3>Heat</h3>

We know that, for an ideal gas, the energy is:

E= c_v n R T

where c_v its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

\Delta E = \Delta Q - \Delta W

where \Delta W is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

\Delta E = 0

\Delta Q = \Delta W

7 0
2 years ago
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