If electromagnetic radiation acted like particles in the double-slit experiment, we would observe one bright band would appear in the center of the screen.
<h3>Bahavior of particles in double-slit experiment</h3>
In a double-slit experiment, single particles, such as photons, pass one at a time through a screen containing two slits.
The photons behave like wave and the constructive interfernce of the waves of these photons will generate a high amplitude wave seen as a bright band in the center of the screen.
Thus, if electromagnetic radiation acted like particles in the double-slit experiment, we would observe one bright band would appear in the center of the screen.
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Answer:
It would be a square and 2d
Explanation:
Opening the valve allows more of the pressurized material out. If the area is decreased, less of the pressurized material is released, and its force ends up more spread out, reducing the pressure. Opening the valve will increase volume of transfer of your liquid.
Answer:
E = 20.03 J
Explanation:
Given that,
The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,
Voltage, V = 12 V
We need to find the energy delivered to the lightbulb filament during 2.00 s.
The energy delivered is given by :
. ....(1)
As,

As per Ohm's law, V = IR

Using formula (1).

So, the energy delivered to the lightbulb filament is 20.03 J.
Answer:
Explanation:
The speed of the astronaut can be found with the help of law of conservation of momentum .
mv = MV , M is mass of astronaut , m is mass of object thrown , v is velocity of object thrown and V is velocity of astronaut.
Putting the values
77.5 x V = .94 x 12
V = .14554 m /s
This will be the uniform velocity of astronaut.
Distance to be covered = 37.3 m
time taken = distance / velocity
= 37.3 / .14554
= 256.28 s
= 4.27 minutes.