Answer:
4.5 W
Explanation:
Applying,
P = V²/(R₁+R₂).................. Equation 1
Where P = Power, V = Voltage, R₁ and R₂ = values of the two resistor.
From the question,
Given: V = 9.00 V, R₁ = 7.00 Ω, R₂ = 11.00 Ω
Substitute these values into equation 1
P = 9²/(7+11)
P = 81/(18)
P = 4.5 Watt.
Hence the power dessipated by the two resistors is 4.5 watt
If the car moves along the distance it will be 16 of the line graph where is independent of the graph
The core has positive charge<span>, the electrons have negative </span>charge. When you are rubbing<span> the </span>glass rod<span> with the </span>silk cloth<span>, electrons are stripped away from the atoms in the </span>glass<span> and transferred to the </span>silk cloth<span>. This leaves the </span>glass rod<span> with more </span>positive<span> than negative </span>charge<span>, so you get a net </span>positive charge<span>.</span>
<u>The tip of the bulb touches the </u>negative end<u> of the battery</u> parts of the light bulb must be touched for the light bulb to light.
What is negative end?
The grounded connection of a car battery is called a negative terminal, and it is typically identified by a minus (-) sign. Typically, it is black and color-coded. The other terminal is a positive terminal that is typically color-coded red or orange and denoted by a plus (+) sign. The grounded terminal attaches to an object that is grounded on the car, usually a bolt in the chassis. The starter motor, as well as other switches, relays, and accessories, are connected to the positive terminal. The car won't start if the battery isn't properly grounded. These days, the majority of car batteries are packed with a charged acid solution that is very explosive and gaseous. In order to keep the battery's inner plates immersed, distilled water may occasionally need to be supplied to the fill-caps.
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<span>Answer:
True: A, C, E
False: B, D
The following representation of this orbital, 5dx2â’y2, depicted when it is bisected by the xy plane, shows the effect of the radial nodes on the orbital contours.</span>
