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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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A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse
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Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg :  crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight  (W)

W= m*g

W= 90kg*9.8 m/s²

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Friction force : Ff

Ff= μk*N Formula (2)   

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax    , ax=0

282 N - 309.58 N = 90*a  

a=  (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

v_{f} = 0.51 \frac{m}{s}

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