Answer:
theroy of plate tectonics
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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By definition, the density of an object is given by:
Where,
M: mass of the object
V: volume of the object
Since the mass and volume of an object are numerical values greater than zero, then it follows that:
It is important to respect the units of each measure.
For this case we can use the grams for the mass and cubic centimeters for the volume.
Answer:
A possible value for density is given by:
Efficiency is defined as the measure of the amount of work or energy is conserved in a certain process. At all times, in every process, work or energy is always lost or wasted due to certain interference. Not all work given is converted to useful work or energy. Thus , efficiency is calculated by dividing the energy or work output to the energy or work input then the value is multiplied by 100 to express efficiency as percentage.
Efficiency = work output / work input
Efficiency = (1020 J / 1200 J) = 85%
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