Answer: Constructive interference
Explanation: Just took the test
The circuit change when the wire is added will see a short circuit occur and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit. Option D. This is further explained below.
<h3>
How does the circuit change when the wire is added?</h3>
Generally, Electronic circuits consist of a series of interconnected parts that form a closed loop through which electricity may flow.
In conclusion, If two wires are linked together, a short circuit will develop, cutting power to bulbs 1 and 2. But there is no impact on bulbs 3 and 4. There is no problem with bulbs 3 and 4.
Read more about circuit
brainly.com/question/21505732
#SPJ1
<h2>5.3 km</h2>
Explanation:
This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.
Let us denote each of the individual displacements by a vector. Consider the unit vectors
as the unit vectors in the direction of East and North respectively.
By simple calculations, we can derive the unit vectors
in the directions North,
South of West and
North of West respectively.
So Total displacement vector = Sum of individual displacement vectors.
Displacement vector = 
Magnitude of Displacement = 
∴ Total displacement = 
Answer:
1.925 μC
Explanation:
Charge: This can be defined as the product of the capacitance of a capacitor and the voltage. The S.I unit of charge is Coulombs (C)
The formula for the charge stored in a capacitor is given as,
Q = CV ................... Equation 1
Where Q = charge, C = Capacitor, V = Voltage.
Note: 1 μF = 10⁻⁶ F
Given: C = 0.55 μF = 0.55×10⁻⁶ F, V = 3.5 V.
Substitute into equation 1
Q = 0.55×10⁻⁶×3.5
Q = 1.925×10⁻⁶ C.
Q = 1.925 μC
Hence the charge on the plate = 1.925 μC