The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs. In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases. In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.
<h3>What is power of the circuit?</h3>
The power of the bulb or any resistor is equal to the product of voltage and current flowing through it.
P = VI
Circuit A has bulbs in series while the circuit B has bulbs in parallel.
When bulb 3 added to circuit A, the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.
The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.
Thus, the last option is correct.
Learn more about power.
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Answer:

Explanation:
We know,
..............(1)
where,
η = Efficiency of the engine
T₁ = Initial Temperature
T₂ = Final Temperature
Q₁ = Heat available initially
Q₂ = Heat after reaching the temperature T₂
Given:
η =0.280
T₁ = 3.50×10² °C = 350°C = 350+273 = 623K
Q₁ = 3.78 × 10³ J
Substituting the values in the equation (1) we get

or

or

⇒ 
Now,
The entropy change (
) is given as:

or

substituting the values in the above equation we get


Answer:

Explanation:
It is given that,
The coordinates of a particle in the metric xy-plane are differentiable functions of time t are given by :


Let D is the distance from the origin. It is given by :

Differentiate above equation wrt t as:

.............(1)
The points are given as, (12,5). Calculating D from these points as :

Put all values in equation (1) as :


So, the particle is moving away from the origin at the rate of 7.076 m/s. Hence, this is the required solution.
The number of times something happens in one second is its "frequency".