Which one of the following terms is used to describe the bending of waves and subsequent spreading around obstacles or the edges
of an opening?
A) refractionB) diffractionC) interferenceD) coherenceE) sonoluminescence
A) refractionB) diffractionC) interferenceD) coherenceE) sonoluminescence
1 answer:
Answer:
B)Diffraction
Explanation:
The concept of diffraction in the field of physics is defined as the deflection of a wave when it crosses an opening or hits the edge of an opaque element. Diffraction is a phenomenon that involves all waves: electromagnetic, radio, sound, etc., and it is possible to predict their behavior using different mathematical approaches. There is a method of analysis called the Huygens principle, which allows us to understand diffraction as a wavefront that is seen as a series of emitters capable of redirecting the wave when it oscillates and thus promotes propagation. Although the waves produced by the oscillators are spherical, their interference causes a flat wave that moves in the same direction as the initial one.
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By definition the friction force is defined as
Where,
Frictional Constant
N = Normal Force -> mg
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E = F*v
Where,
Force
v = Velocity.
Then replacing with our values we have that,
F = 0.5 *34
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E = f v
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The amount of charge enters (and leaves) is 12 Coulomb.
<h3 /><h3>How can we find the value of the amount of charge enters (and leaves)?</h3>Here we are using the formula,
and
We are given,
P = Power of the light bulb = 25 Watt.
V = The amount of voltage where the light bulb operated = 120 Volt.
t= The time when charge enters (and leaves) the light bulb = 1.0 min = = 60 S
We have to find,
I= The amount of current flows in the light bulb.
Q= The amount of charge enters (and leaves) the light bulb.
Now we substitute the values of known parameters in the first equation, we find that,
Or,
The amount of current flows in the light bulb is 0.20 Ampere.
Now, we put the value of I in the second equation,
Or,
So, from the above calculations we can see that the amount of charge enters (and leaves) the light bulb is 12 Coulomb.
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