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expeople1 [14]
2 years ago
5

I need help. which interaction does not take place due to field forces

Physics
1 answer:
Advocard [28]2 years ago
7 0

Answer:

Explanation:

Answer D. Interaction between a pen and paper while you write

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Th answer is "electric attraction is a force that can act at a distance."
Tasya [4]
Electric forces is not action-by-distance. Charged particle emits a electric field radially outwards. It corresponds by the inverse-square, meaning it is 1/r^2.
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3 years ago
WHY does an acceleration cause velocity to increase
harkovskaia [24]

Answer:

When the velocity of an object changes it is said to be accelerating. Acceleration is the rate of change of velocity with time. ... Acceleration occurs anytime an object's speed increases or decreases, or it changes direction.

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3 years ago
Read 2 more answers
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

7 0
3 years ago
Ian walks 2 km to his best friends house, then walks 0.5 km to the library. He then makes 2.5 km walk home. The entire walk took
allochka39001 [22]
1. 2+0.5+2.5= 3. 2km/hr average




2. 14-6=4seconds. 8m/s in 4s = 2m/s acceleration


3. 15m/s divided by 2.5 = 6m/s acceleration
5 0
3 years ago
What type of nuclear radiation is emitted when carbon-14 decays
dalvyx [7]

Answer:

Beta radiation

Explanation:

Beta radiation is a radioactive phenomenon of nuclear decay in which an unstable atom or isotop, by transforming a neutron into a proton, or by transforming a proton into a neutron, becomes stable. For example, the decay of carbon 14 produces beta radiation.

6 0
3 years ago
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