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
It depends, You have to have the length and the width of the crest wave.
Transverse, I think. I may be wrong.
Answer:
Explanation:
By Einstein's Equation of photoelectric effect we know that
here we know that
= energy of the photons incident on the metal
= minimum energy required to remove photons from metal
= kinetic energy of the electrons ejected out of the plate
now we know that it requires 351 nm wavelength of photons to just eject out the electrons
so we can say
here we know that
now we have
now by energy equation above when photon of 303 nm incident on the surface