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zavuch27 [327]
4 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]4 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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Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is
saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

\dot{m_{i}} = \rho_{i} v_{i}A_{i}

Where:

v_{i}: is the initial velocity = 20 m/s

A_{i}: is the inlet area of the nozzle = 60 cm²  

\rho_{i}: is the density of entrance = 2.21 kg/m³

\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!                                                                  

5 0
3 years ago
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An fm radio station broadcasts electromagnetic radiation at a frequency of 100.6 mhz. the wavelength of this radiation is ______
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The wavelength would be 2.980044314115. Reduced it would be 2.980 or just 2.98.
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4 years ago
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Why are SI units used for scientific works ?​
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Energy can be one answer! There are many, but energy is a main one.
5 0
3 years ago
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A moving particle is subject to conservative forces only. When its kinetic energy decreases by 10 J, what happens to its mechani
satela [25.4K]

Answer:

Its mechanical energy is the same.

Explanation:

If forces are only conservative, the mechanical energy will be the same.

It can be different if energy get transformed in another kind of energy like elastic energy for example, although the amount of energy is always the same.

If we just have mechanical energy not geting transformed we have:

Em=K+U

Em: Mechanical energy

K: Kinetic energý

U: Potential energy

Then if Kinetic energy decreases 10J, Potential energy will grow up 10J to keep the same amount of mechanical energy.

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3 years ago
Induced EMF and Current in a Shrinking LoopShrinking Loop. A circular loop of flexible iron wire has an initial circumference of
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Answer:

Explanation:

Let c be the circumference and r be the radius

c = 2πr , r = c / 2π , area A = π r² = π (c/2π )²  = (1/4π) x c²

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c = 161 - 13 x 8 = 57 cm , dc/dt = 13 cm/s  

e = dψ/dt = (1 / 2π )x 57 x 13 x 10⁻⁴ = 118 x 10⁻⁴ V.

4 0
3 years ago
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