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3 years ago

A thermos bottle has two layers, a glass layer and a vacuum layer.

Physics

1 answer:

SIZIF [17.4K]3 years ago

4 0

The vacuum layer is the better thermal insulator

Explanation:

Depending on how they conduct heat, materials can be classified into two types:

Thermal conductos: these are materials that are able to transfer heat efficiently. Examples of thermal conductors are metals in general: in fact, when you heat one side of a metal bar, the other end becomes hot very fast; this means that the heat has been transferred very quickly across the metal rod
Thermal insulators: these are materials that do not transfer heat well. In general, gases or rarefied substances are better insulators, because the particles are more spread apart, and therefore the heat (which is transferred by conduction through collisions between molecules) is transferred less efficiently, due to the large distance between the particles.

In this sense, vacuum is the best possible insulator. This is because vacuum contains no particles at all, so there cannot be transfer of heat by conduction (because there cannot be collisions between molecules), and therefore, vacuum is the best thermal insulator.

Learn more about heat:

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If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol

xenn [34]

Answer:

Average density of Sun is 1.3927 $\frac{g}{cm}$ .

Given:

Radius of Sun = 7.001 × $10^{5}$ km = 7.001 × $10^{10}$ cm

Mass of Sun = 2 × $10^{30}$ kg = 2 × $10^{33}$ g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Solution:

Density of Sun is given by,

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Volume of Sun = $\frac{4}{3} \pi r^{3}$

Volume of Sun = $\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}$

Volume of Sun = 1.436 × $10^{33}$ $cm^{3}$

Density of Sun = $\frac{ 2\times 10^{33} }{1.436 \times 10^{33} }$

Density of Sun = 1.3927 $\frac{g}{cm}$

Thus, Average density of Sun is 1.3927 $\frac{g}{cm}$ .

4 0

2 years ago

A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The

lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s

Explanation:

1) We can find the speed of the object by conservation of energy:

$E_{i} = E_{f}$

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s$

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s$

3) When the object is at the equilibrium position, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s$

I hope it helps you!

8 0

2 years ago

Which body could NOT be called a gas giant?

Dafna11 [192]

Sorry I don’t know but good luck and hope you find it

4 0

2 years ago

Read 2 more answers

Which of the following is distance divided by the change in time?

vovikov84 [41]

(distance covered) divided by (time to cover the distance) is SPEED.

(straight-line distance and direction between start-point and end-point) is DISPLACEMENT.

(amount and direction of change in speed) divided by (time for the change) is ACCELERATION.

(amount and direction of change in momentum) divided by (time for the change) OR (mass) times (acceleration) is FORCE.

4 0

3 years ago

A ____________ is the measure of how far the particles in the medium move away from their normal rest position.

Bezzdna [24]

Explanation:

Amplitude is your answer..

Mate..

hope you like it

5 0

2 years ago