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1. The current in a component is reduced uniformly from 100mA to 20 mA over a period of 8.0 s. What is the charge that flows during this time?

The answer given is 480 mC, but my calculations were

I = 100-20 mA = 80mA

Q= It

= 80 x 8.0 =640 mC

What did I do wrong?

2. A child drinks a liquid of density ρ0 and the child is capable of lowering the pressure at the top of the straw by 10%. The acceleration of free fall is g.

What is the maximum length of straw that would enable the child to drink the liquid?

A. ρ0/ 10ρg

B. 9ρ0/ 10ρg

C. ρ0/ ρg

D. 10ρ0/ ρg

Can you help me a little here? Thx!

The answer given is 480 mC, but my calculations were

I = 100-20 mA = 80mA

Q= It

= 80 x 8.0 =640 mC

What did I do wrong?

2. A child drinks a liquid of density ρ0 and the child is capable of lowering the pressure at the top of the straw by 10%. The acceleration of free fall is g.

What is the maximum length of straw that would enable the child to drink the liquid?

A. ρ0/ 10ρg

B. 9ρ0/ 10ρg

C. ρ0/ ρg

D. 10ρ0/ ρg

Can you help me a little here? Thx!

consider the forces acting on the column of liquid. As it is in eqilibrium these forces must balance.

Upwards there is just atmospheric pressure x the area of the column (as $p=\frac{F}{A}$)

downwards there is 0.9 x this (as it is reduced by 10%) plus the weight of the liquid

the weight of the liquid is therefore equal to 0.1 x the force due to atmospheric pressure, p, so....

$mg=0.1pA$

now the mass of the liquid is its density x its volume $m=\rho A L$

so $\rho A L g = 0.1 p A$

cancelling A's and rearranging gives $L = \frac{p}{10 \rho g}$

therefore I reckon the answer is A and there is a typo error. That or my sums are crap (which I'm sure I'll be pounced upon by a pack of wolves if they are)

Upwards there is just atmospheric pressure x the area of the column (as $p=\frac{F}{A}$)

downwards there is 0.9 x this (as it is reduced by 10%) plus the weight of the liquid

the weight of the liquid is therefore equal to 0.1 x the force due to atmospheric pressure, p, so....

$mg=0.1pA$

now the mass of the liquid is its density x its volume $m=\rho A L$

so $\rho A L g = 0.1 p A$

cancelling A's and rearranging gives $L = \frac{p}{10 \rho g}$

therefore I reckon the answer is A and there is a typo error. That or my sums are crap (which I'm sure I'll be pounced upon by a pack of wolves if they are)

I just did the second one too using Bernoulli's equation and got the same answer as Drummy, he got the answer in before me though. Anyway at least this confirms it, and the typo.

Dharma

How does straw drinking actually work?

ApeXaviour

You reduce the pressure in the straw. Now the atmospheric pressure on the surface of the liquid (outside the straw) is pushing down more than the pressure inside, forcing liquid up the straw. It's like a balanced seesaw, if you decrease the weight on the left side, the right side will push the left side up.

So the liquid then has an upward force (not resultant as yet) of magnitude equal to atmospheric pressure multiplied by cross-sectional area of the straw when it is being sucked?

I've also got two other questions about the method used to obtain the answer for the straw question.

•

Why is it that we assume the liquid is in equilibrium? I would have thought that there could be a possibility that it was accelerating as it went up.

•

What does the L physically represent - is it the maximum length of liquid, and therefore the maximum length of straw?

if the child is capable of lowering the pressure above the liquid by 10% then L is the maximum height of the liquid in the straw.

I imagine this must be a very long straw and the little tike is sucking for all he or she is worth and the liquid is rising up the straw to this height. Not quite enough however to have a drink so the liquid is suspended in equilibrium in the straw. The liquid would no doubt accelerate as it reaches this maximum height but this is irrelevant in this case.

You could think of L as the maximum length of the straw which would enable the child to have a drink. If it would longer than L then the child would die of thirst (worst case scenario).

I used to do a fun trick with classes (when this was on the syllabus) of betting that they could not suck water from a glass on the floor through a plastic tube while they were stood on a table above the glass. They would have to suck the liquid up several metres and its a lot harder than you might think. The theoretical maximum is about 26 ft of water (if my memory serves me). Above this a perfect vaccuum would form above the liquid.

I imagine this must be a very long straw and the little tike is sucking for all he or she is worth and the liquid is rising up the straw to this height. Not quite enough however to have a drink so the liquid is suspended in equilibrium in the straw. The liquid would no doubt accelerate as it reaches this maximum height but this is irrelevant in this case.

You could think of L as the maximum length of the straw which would enable the child to have a drink. If it would longer than L then the child would die of thirst (worst case scenario).

I used to do a fun trick with classes (when this was on the syllabus) of betting that they could not suck water from a glass on the floor through a plastic tube while they were stood on a table above the glass. They would have to suck the liquid up several metres and its a lot harder than you might think. The theoretical maximum is about 26 ft of water (if my memory serves me). Above this a perfect vaccuum would form above the liquid.

Drummy

if the child is capable of lowering the pressure above the liquid by 10% then L is the maximum height of the liquid in the straw.

[indent]As the height of the liquid increases, its weight increases. Therefore, you would need a greater reduction in pressure from the child sucking to provide the force needed to do work against the liquid's weight and the pressure. But because the child can only reduce the pressure by 10%, there is a limit as to how much liquid can be drawn.[/indent]

?

Drummy

I imagine this must be a very long straw and the little tike is sucking for all he or she is worth and the liquid is rising up the straw to this height. Not quite enough however to have a drink so the liquid is suspended in equilibrium in the straw. The liquid would no doubt accelerate as it reaches this maximum height but this is irrelevant in this case.

a) yes, exactly

and

b) I reckon so. (although if the length of the straw were greater than L then the liquid would be suspended at some point below the top of the straw) If the child were succesful in sucking up the liquid then the 10% reduction would be enough to overcome the weight of the liquid. It would then become a much more complicated problem involving fluid flow, fluid friction etc.. I doubt that is intended in this case.

and

b) I reckon so. (although if the length of the straw were greater than L then the liquid would be suspended at some point below the top of the straw) If the child were succesful in sucking up the liquid then the 10% reduction would be enough to overcome the weight of the liquid. It would then become a much more complicated problem involving fluid flow, fluid friction etc.. I doubt that is intended in this case.

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