Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N
Answer:
I'm not sure
Explanation:
I have had that question to Uchida c r go crew in to go be
Energy slowly leaks outward through the radiative diffusion of photons that repeatedly bounce off ions and electrons.
<h3>What is radiative diffusion?</h3>
A radiation zone is a layer of a star's core where energy is mostly carried toward the outside by radiative diffusion and thermal conduction rather than convection.
As photons, energy passes through the radiation zone as electromagnetic radiation.
The radiative diffusion of photons that repeatedly bounce off ions and electrons progressively drains energy outward.
Hence,radiative diffusion is correct answer.
To learn more about radiative diffusion refer:
brainly.com/question/3598352
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