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dolphi86 [110]
10 months ago
14

(fill in the blank) , itself a silver shape beneath the steadfast constellations, simon's dead body moved out toward the open se

a.
Physics
1 answer:
slamgirl [31]10 months ago
7 0

<u>Softly, surrounded by a fringe of inquisitive bright creatures</u>, itself a silver shape beneath the steadfast constellations, Simon's dead body moved out toward the open sea.

This excerpt is from 'Lord of the Flies', by William Golding. It is inspired by the following concepts:

  • adolescent cruelty tendencies
  • his tenure as a teacher at a school for males only 
  • his military service
  • the Coral Island

According to Golding, the subject of the book is that the community is imperfect because humankind is flawed. Considering this, we know that the "flies" in the novel stand for death.

Learn more about the Lord of the Flies here:

brainly.com/question/797646

#SPJ4

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The resistivity of gold is 2.44 × 10-8 Ω · m at room temperature. A gold wire that is 1.8 mm in diameter and 11 cm long carries
tester [92]
Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:

Resistance  =  2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 <span>Ω

P = I^2R = .170^2 (</span>1.055x10^-3 ) = 3.048x10^-5 W
6 0
3 years ago
A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge
Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is \lambda and the net charge per unit length is 2 \lambda.

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let \lambda_i\ and\ \lambda_o are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

\phi=\dfrac{q_{enclosed}}{\epsilon_o}

E.A=\dfrac{q_{enclosed}}{\epsilon_o}

0=\dfrac{\lambda_i+\lambda}{\epsilon_o}

\lambda_i=-\lambda  

For outer surface,

\lambda_i+\lambda_o=2\lambda

-\lambda+\lambda_o=2\lambda

\lambda_o=3\lambda

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}

E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}

Hence, this is the required solution.

6 0
3 years ago
The derived unit for density are g/cm3.<br><br> True<br> False
anzhelika [568]
True, the measurement shown is a derived unit.
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2 years ago
Would u rather/ be nba young boy or polo g
denis-greek [22]

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nba young bruuhh

Explanation:

have a great day and plz mark brainliest!

6 0
2 years ago
Thiết bị nào sau đây không phải là nguồn điện
kumpel [21]

Thiết bị không phải nguồn điện:

Đáp án D

7 0
3 years ago
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