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pantera1 [17]
3 years ago
10

Why would knowing the characteristics of circuits be important in designing electrical circuits?

Physics
2 answers:
zubka84 [21]3 years ago
8 0
Well if you didn't you could make mistakes, which would lead ,in the best case, at a fail of the circuit , or if it goes out of control it could be dangerous

for example you have to know that the wires become hot and they loose their abbilitys as connecters(the hotter it will, the more energy you lose becouse the R will be bigger)
Pachacha [2.7K]3 years ago
4 0
In designing electrical circuits, it is important to know the characteristics of circuits in order to know how to solve related problems that will<span> arise whenever there is certain parameter required.

I hope my answer has come to your help. God bless and have a nice day ahead!
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Masteriza [31]

Answer:

D, the lithosphere. (CRUST AND UPPER MANTLE)

Explanation:

A tectonic plate (also called lithospheric plate) is a massive, irregularly shaped slab of solid rock, generally composed of both continental and oceanic lithosphere. Plate size can vary greatly, from a few hundred to thousands of kilometers across; the Pacific and Antarctic Plates are among the largest. Plate thickness also varies greatly, ranging from less than 15 km for young oceanic lithosphere to about 200 km or more for ancient continental lithosphere (for example, the interior parts of North and South America).

Information found on:

<u>https://pubs.usgs.gov/gip/dynamic/tectonic.html#:~:text=A%20tectonic%20plate%20(also%20called,both%20continental%20and%20oceanic%20lithosphere.&text=Continental%20crust%20is%20composed%20of,such%20as%20quartz%20and%20feldspar.</u>

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3 years ago
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You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
Anarel [89]

Answer:

R = 24.3 m

Explanation:

As we know that the orbital speed is given as

v = \sqrt{\frac{GM}{R + h}}

here we know that

v = 5500 m/s

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now we have

5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}

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now acceleration due to gravity of planet is given as

a = \frac{GM}{R^2}

a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}

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now range of the projectile on the surface of planet is given as

R = \frac{v^2 sin2\theta}{g}

R = \frac{14.6^2 sin(2\times 30.8)}{7.7}

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3 years ago
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Explanation:

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Answer:

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Explanation:

see the above attachment to solve the question and get the answer.

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