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luda_lava [24]
3 years ago
11

A tin can collapses if all air inside it is taken out why

Physics
1 answer:
Veseljchak [2.6K]3 years ago
4 0

That only happens when the tin can is IN air.

In the familiar, comfy part of Earth's atmosphere where we live, the normal pressure of air is around 14.6 pounds on every square inch of everything. That's a big part of the reason why we're built with bodies that generate that same amount of pressure on the INSIDE pressing OUT. That way, we always have the same pressure pushing in both directions, so we know that we won't get crushed or blow up like balloons.

But we have to be careful with our bodies or other things when they're in places where the atmospheric pressure on the outside is NOT normal.

-- When a deep-sea diver goes hundreds of feet down in the ocean, and the pressure of the water is much GREATER than normal air.

-- When an astronaut has to go outside ... where there's NO air ... and fix something on the International Space Station.

When the pressure on the outside becomes very unusual, we have to wear special suits to protect our bodies from the unusual conditions.

The tin can in the story is a lot like our bodies. As long as it has air inside and air outside, the pressure is the same in both directions, so there's no particular force trying to deform the can. But ...

-- If you seal the can with the air inside it, take the can into a vacuum chamber, and pump the air out of the vacuum chamber, then the can only has pressure inside. It'll expand, and eventually spring a little hole in the metal, and all the air inside will blow out.

-- If you take all the air OUT of the can (so the can is REALLY 'empty'), then the pressure on it is all from the outside. In that situation, the can simply collapses, because there's nothing inside to provide pressure in the outward direction.

One more little thing to think about:

When you want some toothpaste to come drizzling out of the tube onto your brush, what do you do ? Do you perhaps squeeze the tube, and increase the pressure on the outside ?

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Introduction to Symbolic Answers, Part B, Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.
lesantik [10]
2cos2(o)-1    is the answer
4 0
4 years ago
Why do clothes often cling together after tumbling in a clothes dryer?
nadya68 [22]

Answer:

Explained

Explanation:

The electrons get rubbed off some items onto others. This causes an excess or electron on one item and deficiency of electron on the other. Therefore an electrostatic force of attraction is produced and hence clothes often cling together after tumbling in a clothes dryer

8 0
3 years ago
A car drives 23m/s east for 35 seconds. What is the displacement?
Lina20 [59]

Answer:

805m

Explanation:

Speed = displacement/time

Speed = 23m/s

Time = 35s

Displacement = speed × time

= 23 × 35

= 805m

5 0
3 years ago
A mixture of nitrogen and xenon gases, at a total pressure of 836 mm Hg, contains 2.80 grams of nitrogen and 24.9 grams of xenon
larisa86 [58]

Answer: Partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

p_A=p_T\times \chi_A

where,

p_A = partial pressure of substance A

p_T = total pressure

\chi_A = mole fraction of substance A

We are given:

m_{N_2}=2.80g

m_{Xe}=24.9g

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

And,

n_A=\frac{m_A}{M_A}

Mole fraction of nitrogen is given as:

\chi_{N_2}=\frac{\frac{m_{N_2}}{M_{N_2}}}{(\frac{m_{N_2}}{M_{N_2}}+\frac{m_{Xe}}{M_{Xe}})}

Molar mass of N_2 = 28 g/mol

Molar mass of Xe =  g/mol

Putting values in above equation, we get:

\chi_{N_2}=\frac{\frac{2.80}{28}}{\frac{2.80}{28}+\frac{24.9}{131}}

\chi_{N_2}=\frac{0.100}{0.100+0.190}=0.345

To calculate the mole fraction of xenon, we use the equation:

\chi_{Xe}+\chi_{N_2}=1\\\\\chi_{Xe}=1-0.345=0.655

p_{N_2}=836mmHg\times 0.345=288mmHg

p_{Xe}=836mmHg\times 0.655=548mmHg

Thus partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

6 0
4 years ago
Calculate the pressure exerted on the heel of a boy’s foot if the boy weighs 80 N and he lands on one heel,which has an area of
cricket20 [7]

Pressure at a given surface is given as ratio of normal force and area

so here force due to heel of the shoes is given as 80 N

and the area of the heel is given as 16 cm^2

so we can say

P = \frac{F}{A}

here we have

F = 80 N

A = 16 cm^2 = 16 * 10^{-4} m^2

P = \frac{80}{16 * 10^{-4}}

P = 5 * 10^4 N/m^2

so pressure at the surface due to its heel will be 5 * 10^4 N/m^2

3 0
3 years ago
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