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weqwewe [10]
3 years ago
5

Fiberglass, an insulator, can be found in the wals and roofs of some houses and buildings. Why would an insulator be needed insi

de of buildings in the winter? It makes the building look pretty B) It keeps the heat inside of the building. C)It keeps the heat outside of the building. D) It keeps the cool air inside of the building. A) A it makes the building look pretty
Physics
2 answers:
GREYUIT [131]3 years ago
5 0

It keeps the heat inside of the building

Fudgin [204]3 years ago
4 0

Answer:

B) It keeps the heat inside of the building.

Explanation:

An insulator during winter is needed to stop the heat flow from going outside.

A Fiberglass insulation is made of plastic and tiny pieces of glass. Glass is a non conductive heat material that will stop the scape of heat from a building during winter. In winter if one place is warmer than other, for example the inside of a building and the outside respectively,  heat will tend to flow from the warmest place to the coolest place in order to valance the difference in temperature and reach a thermal equilibrium, to prevent this we use insulators.

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What cultular value does this melodic line portray?<br>a.jubilant<br>b.reserved<br>c.rich<br>d.poor​
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<em>What cultular value does this melodic line portray?</em>

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6 0
3 years ago
An asteroid has acquired a net negative charge of 149 C from being bombarded by the solar wind over the years, and is currently
bearhunter [10]

Answer:

93.125 × 10^(19)

Explanation:

We are told the asteroid has acquired a net negative charge of 149 C.

Thus;

Q = -149 C

charge on electron has a value of:

e = -1.6 × 10^(-19) C

Now, for us to determine the excess electrons on the asteroid, we will just divide the net charge in excess on the asteroid by the charge of a single electron.

Thus;

n = Q/e

n = -149/(-1.6 × 10^(-19))

n = 93.125 × 10^(19)

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5 0
3 years ago
(a) Calculate the magnitude of the gravitational force exerted on a 445-kg satellite that is a distance of 1.77 earth radii from
victus00 [196]

Answer:

a)1396.52 N

b)1396.52 N

c)a_{satellite}= 3.13 m/sec^2

d)a_{earth}=2.32\times10^{-22} m/s^2

Explanation:

The force experienced by the satelite is giveb by

F= \frac{Gm_{satellite}m_{earth}}{r^2} \\

m_{satellite}= 445 Kg

m_{earth}= 6×10^24 Kg

radius r= 1.77Re= 1.77×6.38×10^6 m

now putting values we get

F= \frac{6.67\times10^{-11}(445)(6\times10^24)}{(1.77\times6.38\times10^6)^2}

⇒F= 1396.52 N

now,

a_{satellite}= \frac{F}{m_{satellite}}

a_{satellite}= \frac{1396.52}{445}

a_{satellite}= 3.13 m/sec^2

also,

a_{earth}= \frac{F}{m_{earth}}

a_{earth}= \frac{1396.52}{(6\times10^24)}

a_{earth}=2.32\times10^{-22} m/s^2

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