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djyliett [7]
3 years ago
13

4. Provide an example of an isolated energy system and explain how it could be changed to create an open energy system. You may

use the examples provided in the lesson or another example.
Physics
2 answers:
levacccp [35]3 years ago
5 0

Answer:

Explanation:

Isolated energy system does not allow the energy or matter transfer, whereas, Open energy system allows energy and matter to exchange. For example, Insulated container (Thermos bottle), if the insulated bottle is filled with hot tea and the container is closed then there will be no exchange of matter and energy but as we open up the top of container then the heat will escape to the surrounding and exchange of matter and energy will take place.

GarryVolchara [31]3 years ago
4 0
<span>Easy, take the top off your Thermos bottle filled with hot coffee. Assuming perfect insulation, that hot coffee is isolated from the environment; but when the top is opened the heat can now escape to that environment.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
</span>
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A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

3 0
2 years ago
Definition of AM and Fm Wave? in your own word
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Answer:

In AM broadcasting, the amplitude of the carrier wave is modulated to encode the original sound. In FM broadcasting, the frequency of the carrier wave is modulated to encode the sound. A radio receiver extracts the original program sound from the modulated radio signal and reproduces the sound in a loudspeaker.

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3 years ago
an op amp in unity gain configuration (buffer) with slew rate of 5v/us is used to amplify a sinusoidal signal with a frequency o
ZanzabumX [31]

Answer:

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Therefore, from the above equation we can write,

&&5 \frac{V}{\mu s} = 2 \pi 100 kHz \times V_{max}\\&or,& V_{max} = \frac{5V}{10^{-6} s \times 2 \pi 100 \times 10^{3} Hz}\\&or,& V_{max} = \frac{5}{2 \pi \times 10^{-1}} V = 7.96 V

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Answer:

an ellipse with the Sun at one focus  or D

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