Complete question:
A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from the bulb, the amplitude of the electric field is 3.78 V/m. What is the average intensity of the light?
Answer:
The average intensity of the light is 0.02 W/m²
Explanation:
Given;
Amplitude of the electric field, E₀ = 3.78 V/m
The average intensity of the light is calculated as follows;

where;
is the average intensity of the light
c is speed of light = 3 x 10⁸ m/s

Therefore, the average intensity of the light is 0.02 W/m²
If the three spoon touch nothing happens because they are all at room Temperature
Answer:
486,750 kg*m/s
Explanation:
Momentum is mass*velocity
M = m*v
M = 8850kg*55m/s
M = 486,750 kg*m/s
In order for the object to move upward, it needs an upward force
that's at least equal to its own weight.
Weight = (mass) x (gravity) = (35 kg) x (9.8 m/s²) = 343 N.
The engine thrust has to be more than 343 N.
It would be a) 2 electrons