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kvasek [131]
2 years ago
5

Answer pls urgent pls​

Physics
2 answers:
Simora [160]2 years ago
6 0

Answer:

\huge\boxed{\sf brittleness}

Explanation:

Iron is a metal which have luster (shine) so it is lustrous (shiny).

Iron is sonorous. When it is hit with a hard object, it produces ringing sounds.

Iron is not brittle. It cannot be easily broken. So, Iron does not show brittleness (It cannot be easily broken)

\rule[225]{225}{2}

Hope this helped!

<h3>~AnonymousHelper1807</h3>
Drupady [299]2 years ago
4 0

Answer:

Brittleness

Explanation:

Lustrous means shiny

sonorous means capable of producing a deep or ringing sound

and iron isn't brittle or weak

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Thermal energy, radiant energy
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3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

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nirvana33 [79]
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IgorLugansk [536]

Answer:

trigonometry (guessing)

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ellipse: is the shape of an orbit : looks like an oval

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https://science.howstuffworks.com/question224.htm

By comparing the intrinsic brightness to the star's apparent brightness we can calculate the distance of stars

1/r^2 rule states that the apparent brightness of a light source is proportional to the square of its distance.Jan 11, 2022

https://www.space.com/30417-parallax.html

alternative distance measurement for stars used by most astronomers is the parsec. A star with a parallax angle of 1 arcsecond has a distance of 1 parsec, or 1 parsec per arcsecond of parallax, which is about 3.26 light years

blossoms.mit.edu

.

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Why don’t metals break when hit with hammer
Verizon [17]

Because metallic bonds involve all of the metal atoms in a piece of metal sharing all of their valence electrons with "delocalized" bonds.

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