Interference and diffraction are the phenomena that support only the wave theory of light. Options 2 and 3 are correct.
<h3 /><h3>What is the interference of waves?</h3>
The result of two or more wave trains flowing in opposite directions on a crossing or coinciding pathways. This phenomenon is known as the interference of waves.
The phenomenon of interference occurs when two wave pulses are traveling along a string toward each other.
The light wave hypothesis states that light behaves like a wave. Since light is an electromagnetic wave, it may be transmitted without a physical medium.
Light has magnetic and electric fields, much like electromagnetic waves do.
Transverse waves, such as those seen in light waves, oscillate in the same direction as the wave's path. A wave of light may experience interference as well as diffraction as a result of these properties.
All of the remaining options are the light phenomenon.
Hence, options 2 and 3 are correct.
To learn more about the interference of waves refer to the link;
brainly.com/question/16098226
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Answer:
0.5A
Explanation:
Using 
,
R is the resistance (in Ohms)
V is the voltage (in V)
I is the current (in A)
I = 0.5A
Answer:
✓ A cyclone device accumulates fine particulates from the air by making a dirty air stream flow in a spiral path inside a cylindrical chamber.
✘ It consists of several long and narrow fabric filter bags suspended upside-down in a large enclosure.
✓ When dirty air enters the chamber, the larger particulates strike the chamber wall and fall into a conical dust hopper at the bottom.
✘ Fans blow dirt-filled air upward from the bottom of the enclosure, trapping dirt particles inside the filter bags and releasing clean air from the top.
✓ The top of the chamber has an outlet that lets out cleaned air.
Basically, any of these choices that have the word "filter" are wrong. The point of the cyclone device is to separate the particles without the use of filters. You can tell the right answers based on the picture attached below.
The electric force exerted by an electric field of intensity E on a charge q is equal to the product between E and q, so:
<span>1) The differential equation that models the RC circuit is :
(d/dt)V_capacitor </span>+ (V_capacitor/RC) = (V_source/<span>RC)</span>
<span>Where the time constant of the circuit is defined by the product of R*C
Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s
2) C<span>harge of the capacitor 1.57 time constants
1.57*(2.742) = 4.3048 s
The solution of the differential equation is
</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T) + </span>V_capac(∞)
Since the capacitor is initially uncharged V_capac(0) = 0
And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V
This means,
V_capac (t) = (-9V)e ^(-t /T) + 9V
The charge in a capacitor is defined as Q = C*V
Where C is the capacitance and V is the Voltage across
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V
V_capac (4.3048 s) = 7.1275 V
Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C
3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V
Q (∞) = 89.9mF*(9V) = 0.8091 C
