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Nady [450]
3 years ago
12

The tin can with water in its bottom is heated to boil water and the steam is allowed to escape for some time. The open mouth is

sealed with an air-tight cap and cooled under tap water. The tin can get crushed, why?
Physics
1 answer:
Norma-Jean [14]3 years ago
6 0

Explanation:

Water does expand with heat (and contract with cooling), but the amount of expansion is pretty small. So when you boil a can filled with water and seal it, the water will contract slightly as it cools. The can may kink slightly, but that will be it. Actually, most likely the only things you will be able to see is then top and bottom will be sucked in and go concave. Just like a commercial can of beans.

Now if you have a can with a little water and a big air space, things are completely different.

As the water boils, water vapour is given off. Steam. Let it boils for a minute just to make sure (nearly) all the air is expelled and the can is filled with steam.

Now when you put the lid on and cool the can, that steam condenses back to water, and goes from filling the can to a few drops of water. The can is now filled (if that is the right word) with a near vacuum, The air pressure, 15 lbs/square inch, will be pressing on every surface of the can, with nothing inside the can to resist it.

The can will crumple before your eyes.

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The most likely answer to this problem would be (1) more mass and more inertia.

A 15-kilogram cart at rest and a 5-kilogram box would make up a 20-kilogram cart and box that is at rest on a horizontal surface. The mass changed into something more, of course, as a result of combining the two object into one and by combining the two objects' mass, the inertia that it previously possessed as a cart by itself was increased when the inertia of the box was also combined to the cart.
6 0
2 years ago
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
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Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

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(1.05) / (9.5 x 3.262) x (km · LY · psc) / (km · LY) x (10¹⁵⁻¹²) =

(0.03388) x (psc) x (10³) =

33.88 parsecs


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Answer:

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Explanation:

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