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3 years ago

paper by itself catches fire easily whereas a piece of paper wrapped around an alluminium pipe does not.

Physics

1 answer:

Julli [10]3 years ago

7 0

Answer:

The paper does not catch fire when wrapped around aluminium pipe because aluminium absorbs the heat, so paper does not attain its ignition temperature.

Explanation:

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

SIZIF [17.4K]

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:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

4 0

3 years ago

A student holds one end of a thread, which is wrapped around a cylindrical spool, as shown above. The student then drops the spo

lesya [120]

by energy conservation we know that

KE or rotation + KE of translation = gravitational PE

now we have

$\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 = mgH$

also we know that

$v = R\omega$

now we have

$\frac{1}{2}(\frac{1}{2}mR^2)\omega^2 + \frac{1}{2}m(R\omega)^2 = mgH$

$\frac{3}{4}mR^2\omega^2 = mgH$

$\omega = \sqrt{\frac{4gH}{3}}/R$

now when it is rolling on ground the torque acting on it due to friction force is given by

$\tau = R F_f$

$\tau = \mu mg R$

$\alpha = \frac{\mu mg R}{\frac{1}{2}mR^2}$

$\alpha = \frac{2 \mu g}{R}$

now angular speed at any time is given as

$\omega = \omega_i + \alpha t$

$\omega = \sqrt{\frac{4gH}{3}}/R -\frac{2 \mu g}{R} t$

so above is the angular speed in terms of time "t"

7 0

4 years ago

A parallel-plate capacitor is charged and then is disconnected from the battery. By what factor does the stored energy change wh

andrew-mc [135]

Answer:

0.5

Explanation:

The energy of a charged capacitor is given by $\frac{1}{2}QC$ where $Q$ is the charge on it and $C$ is its capacitance. The capacitance is defined, geometrically, as

$\dfrac{\epsilon A}{d}$

where $\epsilon$ is a constant that is determined by the material between the plates, $A$ is the area of the plates and $d$ is the distance between them. It is then that the capacitance is linearly inversely proportional to the separation distance; as the distance increases, the capacitance reduces.

Because it is linear, when the separation distance is doubled, other factors remaining constant, the capacitance is halved. Because the capacitance is halved, the energy is halved.

Let's see the mathematics.

Initial capacitance, $C_1$ at initial separation, $d_1$

$C_1=\dfrac{\epsilon A}{d_1}$

If separation is doubled, separation becomes $d_2=2d_1$ . Then the capacitance becomes

$C_2=\dfrac{\epsilon A}{d_2}$

$C_2=\dfrac{\epsilon A}{2d_1}$

$C_2=0.5\dfrac{\epsilon A}{d_1}$

$C_2=0.5C_1$

Initial energy, $E_1=\frac{1}{2}QC_1$

Final energy, $E_2=\frac{1}{2}QC_2= \frac{1}{2}Q\times0.5C_1$

$E_2=0.5\times\frac{1}{2}QC_1=0.5E_1$

Hence, $\frac{E_2}{E_1}=0.5$

5 0

3 years ago

What does osteoclasts do

Maslowich

A cell that nibbles at and breaks down bone and is responsible for bone resorption.

3 0

3 years ago

Read 2 more answers

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle (fig. 5-34) so that the passeng

almond37 [142]

At top, 2 forces act on passengers: normal force, force of gravity. The two act downward in the direction of the center of the circle.
To answer the problem we would be using this formula which is apply N2L:
Fnet = ma
mg + N = mv^2/r
minimum speed occurs when N = 0
mg = mv^2/r
v = sqrt (rg)
= √94.09
v = 9.7 m/s

4 0

3 years ago

paper by itself catches fire easily whereas a piece of paper wrapped around an alluminium pipe does not.​

paper by itself catches fire easily whereas a piece of paper wrapped around an alluminium pipe does not.