Argon, neon, and xenon are examples of

A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m

belka [17]

Answer:

Energy stored in the capacitor is $U=2.7\times 10^{-5}\ J$

Explanation:

It is given that,

Charge, $q=1.5\ \mu C=1.5\times 10^{-6}\ C$

Potential difference, V = 36 V

We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

$U=\dfrac{1}{2}q\times V$

$U=\dfrac{1}{2}\times 1.5\times 10^{-6}\times 36$

U = 0.000027 J

$U=2.7\times 10^{-5}\ J$

So, the potential energy is stored in the capacitor is $U=2.7\times 10^{-5}\ J$ . Hence, this is the required solution.

4 0

3 years ago

Read 2 more answers

Which is the correct answer ?

hodyreva [135]

D ............................

7 0

3 years ago

Read 2 more answers

An electron is confined to a one dimensional infinite potential well 150 pm. How much energy must it absorb if it is to jump to

igor_vitrenko [27]

Answer:

\Delta E=1.22\times 10^{-22}J

Explanation:

The energy of electron in any state is given by $E=\frac{n^2h^2}{8mL^2}$ here h is planck's constant n is state of electron L is the infinte potential well m is the mass of electron

We know that $h=6.67\times 10^{-34}$

Potential well dimension = $150pm=150\times 10^{-12}m$

Mass of electron $=9.1\times 10^{-31}kg$

So energy required to electron to jump from ground state to 3rd state

$\Delta E=\frac{h^2}{8mL^2}\left ( 3^2-1^2 \right )$

$\Delta E=\frac{\left ( 6.67\times 10^{-34} \right )^2}{8\times 9.1\times 10^{-31}(150\times 10^{-12})^2}\left ( 9-1 \right )$

$\Delta E=1.22\times 10^{-22}J$

7 0

3 years ago

Part C Let’s start the analysis by looking at your “extreme usage” cases. Compare the two cases in detail—low usage period versu

Serga [27]

Answer:

Day 7 DataUsage notes (since last reading)day & datetimekWh readingkWh usedhours elapsedavg. kW usedb.Usage Extremes: Data CollectionFor this experiment, you’ll measure electrical usage during a time period when you expect to havevery light electrical usage (for instance, while you’re asleep at night or during the day when no oneis at home). Likewise you’ll measure electrical usage during a time period when you expect to have heavier than average electrical usage. This time period might be in the evening, when lights and other appliances are on. Both of these time periods should be at least 4 hours long, to increase the accuracy of your results. Record your results in the tables below for each situation. For each time period, you’ll need to takean initial and a final reading.Type your response here:Low Usage - Initial Readingday & datetimekWh readingLow Usage - Final ReadingEnergy Usage Notesday & datetimekWh readingkWh usedhours elapsedavg. kW usedHigh Usage - Initial Readingday & datetimekWh reading4

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3 years ago

Is that statement true or false?

Softa [21]

False, they don’t use the lures to produce light. Instead they use a strong flash of bioluminescence to scare off predators

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3 years ago